Jézusom mennyi... Semmi értelme ennyit feltenni. Mindegyiknél ugyanazt kell csinálni. Behelyettesíteni a megoldó képletbe:
`x_(1,2) = (-b+-sqrt(b^2-4ac))/(2a)`, ahol `a, b, c` a másodfokú együtthatói: `ax^2+bx+c`

1. `9x^2+9x+2 = 0`
`x_(1,2) = (-9+-sqrt(9^2-4*9*2))/(2*9) ``=`` (-9+-sqrt(81-72))/18 ``=`` (-9+-sqrt(9))/18 ``=`` (-3+-1)/6 = {(x_1 = -2/6 = -1/3), (x_2 = -4/6 = -2/3):}`

2. `9x^2+3x-2 = 0`
`x_(1,2) = (-3+-sqrt(3^2-4*9*(-2)))/(2*9) ``=`` (-3+-sqrt(9+72))/18 ``=`` (-3+-sqrt(81))/18 ``=`` (-1+-3)/6 = {(x_1 = 2/6 = 1/3), (x_2 = -4/6 = -2/3):}`

3. `9x^2-9x-2 = 0`
`x_(1,2) = (9+-sqrt((-9)^2-4*9*(-2)))/(2*9) ``=`` (9+-sqrt(81+72))/18 ``=`` (9+-sqrt(153))/18 ``=`` (3+-sqrt 17)/6 = {(x_1 = (3+sqrt 17)/6), (x_2 = (3-sqrt 17)/6):}`

4. `6x^2+5x+1 = 0`
`x_(1,2) = (-5+-sqrt(5^2-4*6*1))/(2*6) ``=`` (-5+-sqrt(25-24))/12 ``=`` (-5+-1)/12 = {(x_1 = -4/12 = -1/3), (x_2 = -6/12 = -1/2):}`

5. `6x^2+x-1 = 0`
`x_(1,2) = (-1+-sqrt(1^2-4*6*(-1)))/(2*6) ``=`` (-1+-sqrt(1+24))/12 ``=`` (-1+-5)/12 = {(x_1 = 4/12 = 1/3), (x_2 = -6/12 = -1/2):}`

6. `6x^2-x-1 = 0`
`x_(1,2) = (1+-sqrt((-1)^2-4*6*(-1)))/(2*6) ``=`` (1+-sqrt(1+24))/12 ``=`` (1+-5)/12 = {(x_1 = 6/12 = 1/2), (x_2 = -4/12 = -1/3):}`

7. `6x^2-5x+1 = 0`
`x_(1,2) = (5+-sqrt((-5)^2-4*6*1))/(2*6) ``=`` (5+-sqrt(25-24))/12 ``=`` (5+-1)/12 = {(x_1 = 6/12 = 1/2), (x_2 = 4/12 = 1/3):}`

8. `5x^2+7x+2 = 0`
`x_(1,2) = (-7+-sqrt(7^2-4*5*2))/(2*5) ``=`` (-7+-sqrt(49-40))/10 ``=`` (-7+-3)/10 = {(x_1 = -4/10 = -2/5), (x_2 = -10/10 = -1):}`

9. `5x^2-7x+2 = 0`
`x_(1,2) = (7+-sqrt((-7)^2-4*5*2))/(2*5) ``=`` (7+-sqrt(49-40))/10 ``=`` (7+-3)/10 = {(x_1 = 10/10 = 1), (x_2 = 4/10 = 2/5):}`

10. `5x^2-3x-2 = 0`
`x_(1,2) = (3+-sqrt((-3)^2-4*5*(-2)))/(2*5) ``=`` (3+-sqrt(9+40))/10 ``=`` (3+-7)/10 = {(x_1 = 10/10 = 1), (x_2 = -4/10 = -2/5):}`

11. `5x^2+3x-2 = 0` (gondolom, az eddigiek alapján)
`x_(1,2) = (-3+-sqrt(3^2-4*5*(-2)))/(2*5) ``=`` (-3+-sqrt(9+40))/10 ``=`` (-3+-7)/10 = {(x_1 = 4/10 = 2/5), (x_2 = -10/10 = -1):}`

12. `5x^2+11x+2 = 0`
`x_(1,2) = (-11+-sqrt(11^2-4*5*2))/(2*5) ``=`` (-11+-sqrt(121-40))/10 ``=`` (-11+-9)/10 = {(x_1 = -2/10 = -1/5), (x_2 = -20/10 = -2):}`

13. `5x^2-9x-2 = 0`
`x_(1,2) = (9+-sqrt((-9)^2-4*5*(-2)))/(2*5) ``=`` (9+-sqrt(81+40))/10 ``=`` (9+-11)/10 = {(x_1 = 20/10 = 2), (x_2 = -2/10 = -1/5):}`

14. `5x^2+9x-2 = 0`
`x_(1,2) = (-9+-sqrt(9^2-4*5*(-2)))/(2*5) ``=`` (-9+-sqrt(81+40))/10 ``=`` (-9+-11)/10 = {(x_1 = 2/10 = 1/5), (x_2 = -20/10 = -2):}`

15. `5x^2-11x+2 = 0`
`x_(1,2) = (11+-sqrt((-11)^2-4*5*2))/(2*5) ``=`` (11+-sqrt(121-40))/10 ``=`` (11+-9)/10 = {(x_1 = 20/10 = 2), (x_2 = 2/10 = 1/5):}`

16. `6x^2-23x+15 = 0`
`x_(1,2) = (23+-sqrt((-23)^2-4*6*15))/(2*6) ``=`` (23+-sqrt(529-360))/12 ``=`` (23+-13)/12 = {(x_1 = 36/12 = 3), (x_2 = 10/12 = 5/6):}`

17. `6x^2+13x-15 = 0`
`x_(1,2) = (-13+-sqrt(13^2-4*6*(-15)))/(2*6) ``=`` (-13+-sqrt(169+360))/12 ``=`` (-13+-23)/12 = {(x_1 = 10/12 = 5/6), (x_2 = -36/12 = -3):}`

18. `6x^2+23x+15 = 0`
`x_(1,2) = (-23+-sqrt(23^2-4*6*15))/(2*6) ``=`` (-23+-sqrt(529-360))/12 ``=`` (-23+-13)/12 = {(x_1 = -10/12 = -5/6), (x_2 = -36/12 = -3):}`

19. `6x^2-13x-15 = 0`
`x_(1,2) = (13+-sqrt((-13)^2-4*6*(-15)))/(2*6) ``=`` (13+-sqrt(169+360))/12 ``=`` (13+-23)/12 = {(x_1 = 36/12 = 3), (x_2 = -10/12 = -5/6):}`

20.
Na végre egy kis változatosság! Itt ugyanúgy lehet használni a megoldó képletet úgy, hogy a hiányzó tag együtthatójához 0-t helyettesítesz be, de meglehet ezt egyszerűbben is oldani!
`x^2-9 = 0 " /"+9`
`x^2 = 9 " /"sqrt`
`x_(1,2) = +-3 = {(x_1 = 3), (x_2 = -3):}`

21. `x^2+6x+9 = 0`
`x_(1,2) = (-6+-sqrt(6^2-4*1*9))/(2*1) ``=`` (-6+-sqrt(36-36))/2 ``=`` (-6+-0)/2 = {(x_1 = -6/2 = -3), (x_2 = -6/2 = -3):}`
(Vagyis itt technikailag 1 gyökünk van, de erre azt mondják, hogy az egy kétszeres gyök.)

22. `x^2-6x+9 = 0`
`x_(1,2) = (6+-sqrt((-6)^2-4*1*9))/(2*1) ``=`` (6+-sqrt(36-36))/2 ``=`` (6+-0)/2 = {(x_1 = 6/2 = 3), (x_2 = 6/2 = 3):}`

23. `x^2+8x+16 = 0`
`x_(1,2) = (-8+-sqrt(8^2-4*1*16))/(2*1) ``=`` (-8+-sqrt(64-64))/2 ``=`` (-8+-0)/2 = {(x_1 = -8/2 = -4), (x_2 = -8/2 = -4):}`

24. `x^2-8x+16 = 0`
`x_(1,2) = (8+-sqrt((-8)^2-4*1*16))/(2*1) ``=`` (8+-sqrt(64-64))/2 ``=`` (8+-0)/2 = {(x_1 = 8/2 = 4), (x_2 = 8/2 = 4):}`

25. `x^2-16 = 0 " /"+16`
`x^2 = 16 " /"sqrt`
`x_(1,2) = +-4 = {(x_1 = 4), (x_2 = -4):}`

26. `2x^2-50 = 0 " /"+50`
`2x^2 = 50 " /":2`
`x^2 = 25 " /"sqrt`
`x_(1,2) = +-5 = {(x_1 = 5), (x_2 = -5):}`

27. `3x^2-243 = 0 " /"+243`
`3x^2 = 243 " /":3`
`x^2 = 81 " /"sqrt`
`x_(1,2) = +-9 = {(x_1 = 9), (x_2 = -9):}`

28. `5x^2+10x+15 = 0`
`x_(1,2) = (-10+-sqrt(10^2-4*5*15))/(2*5) ``=`` (-10+-sqrt(100-300))/10 ``=`` cancel( (-10+-10 sqrt(2)*sqrt(-1))/10 = {(x_1 = -1+sqrt(2)*sqrt(-1)), (x_2 = -1-sqrt(2)*sqrt(-1)):} )`
Ne foglalkozz azzal, amit az áthúzás alatt írtam. Lényeg az, hogy mivel a gyökjel alatt negatív jön ki, ezért nincs megoldása a másodfokúnak! (Legalább is a valós számok halmazán )

29. `5x^2-10x+5 = 0`
`x_(1,2) = (10+-sqrt((-10)^2-4*5*5))/(2*5) ``=`` (10+-sqrt(100-100))/10 ``=`` (10+-0)/10 = {(x_1 = 10/10 = 1), (x_2 = 10/10 = 1):}`

30. `5x^2-5 = 0 " /"+5`
`5x^2 = 5 " /":5`
`x^2 = 1 " /"sqrt`
`x_(1,2) = +-1 = {(x_1 = 1), (x_2 = -1):}`

31. `2x^2-18 = 0 " /"+18`
`2x^2 = 18 " /":2`
`x^2 = 9 " /"sqrt`
`x_(1,2) = +-3 = {(x_1 = 3), (x_2 = -3):}`

32. `10x^2+16x+6 = 0`
`x_(1,2) = (-16+-sqrt(16^2-4*10*6))/(2*10) ``=`` (-16+-sqrt(256-240))/20 ``=`` (-16+-4)/20 = {(x_1 = -12/20 = 3/5), (x_2 = -20/20 = -1):}`