Question

3y² – 20y - 23 = 0 solve by quadratic equation by formula method​

Answer 1

Step-by-step explanation:

By comparing the given quadratic equation with the general form of quadratic equation ax2 + bx + c = 0, we get

a = 3, b = -20 and c = -23

x = [-b ± √b2 - 4ac]/2a

x = [-(-20) ± √(-20)2 - 4(3)(-23)]/2(3)

x = [20 ± √(400 + 276)]/6

x = [20 ± √676]/6

x = [20 ± 26]/6

x = [20 + 26]/6

x = 46/6

x = 23/3

x = [20 - 26]/6

x = -6/6

x = -1

Answer 2

step by step explanation:

3y²-(23-3)y-23 =0

3y²-23y+3y-23=0

y(3y-23) +1(3y-23) =0

(3y-23) (y+1) =0

1) 3y-23=0, 3y=23, y=23/3.

2) y+1, y=-1.