What are the solutions to the equation (5x-4)^2=81 ?

The solutions to the equation (5x-4)^2=81 are x= 13/5 ,x=-1

Find the zeros of (5x-4)^2=81

The zeros of (5x-4)^2=81 are x= 13/5 ,x=-1

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(5 x - 4)^{2} = 81

x = \frac{13}{5}, x = - 1

Decimal Notation

x = 2.6, x = - 1

Solution steps

(5 x - 4)^{2} = 81

For

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

the solutions are

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

Solve

5 x - 4 = 81 : x = \frac{13}{5}

Solve

5 x - 4 = - 81 : x = - 1

The solutions to the quadratic equation are:

x = \frac{13}{5}, x = - 1

4 x^{2} = - 9

x^{2} + 14 x + 18 = 0

(x + 5) (x + 4) = 0

2 (3 x + 5)^{2} - 2 = 70

5 x^{2} - 15 x + 1 = 0

What are the solutions to the equation (5x-4)^2=81 ?
The solutions to the equation (5x-4)^2=81 are x= 13/5 ,x=-1
Find the zeros of (5x-4)^2=81
The zeros of (5x-4)^2=81 are x= 13/5 ,x=-1