Question

solve by completing square
method 3y²-20y-23=0​

Answer 1

Answer:

Comparing 3y2 - 20y - 23 = 0 and ax2 + bx + c = 0, we get

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

Then,

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

Step-by-step explanation:

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