What are the solutions to the equation -5(2x-3)^2=-125 ?

The solutions to the equation -5(2x-3)^2=-125 are x=4,x=-1

Find the zeros of -5(2x-3)^2=-125

The zeros of -5(2x-3)^2=-125 are x=4,x=-1

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- 5 (2 x - 3)^{2} = - 125

x = 4, x = - 1

Solution steps

- 5 (2 x - 3)^{2} = - 125

Divide both sides by

- 5

(2 x - 3)^{2} = 25

For

(g (x))^{2} = f (a)

the solutions are

g (x) = f (a), - f (a)

Solve

2 x - 3 = 25 : x = 4

Solve

2 x - 3 = - 25 : x = - 1

The solutions to the quadratic equation are:

x = 4, x = - 1

x, x^{2} - 8 k x - 9 = 0

x, 5 x^{2} + x y^{2} = C

2 x^{2} + 3 x + 6

x^{2} - x - 20 = - x

(x - 6) (x + 6) - (x - 3) 2 = 0

What are the solutions to the equation -5(2x-3)^2=-125 ?
The solutions to the equation -5(2x-3)^2=-125 are x=4,x=-1
Find the zeros of -5(2x-3)^2=-125
The zeros of -5(2x-3)^2=-125 are x=4,x=-1