What are the solutions to the equation 25(1-4x)^2-28=-12 ?

The solutions to the equation 25(1-4x)^2-28=-12 are x= 1/20 ,x= 9/20

Find the zeros of 25(1-4x)^2-28=-12

The zeros of 25(1-4x)^2-28=-12 are x= 1/20 ,x= 9/20

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25 (1 - 4 x)^{2} - 28 = - 12

x = \frac{1}{20}, x = \frac{9}{20}

Decimal Notation

x = 0.05, x = 0.45

Solution steps

25 (1 - 4 x)^{2} - 28 = - 12

Move

28

to the right side

25 (1 - 4 x)^{2} = 16

Divide both sides by

25

(1 - 4 x)^{2} = \frac{16}{25}

For

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

the solutions are

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

Solve

1 - 4 x = \frac{16}{25} : x = \frac{1}{20}

Solve

1 - 4 x = - \frac{16}{25} : x = \frac{9}{20}

The solutions to the quadratic equation are:

x = \frac{1}{20}, x = \frac{9}{20}

x^{2} - 4 x = 4 x

(x - 3)^{2} = \frac{x ^{2} - 9}{3}

0 = 34 x + 60 - (2 x^{2} + 200)

x, x^{2} + 2 y^{5} = 10 x y

\frac{14}{5} x - \frac{1}{20} x^{2} = 42.5

What are the solutions to the equation 25(1-4x)^2-28=-12 ?
The solutions to the equation 25(1-4x)^2-28=-12 are x= 1/20 ,x= 9/20
Find the zeros of 25(1-4x)^2-28=-12
The zeros of 25(1-4x)^2-28=-12 are x= 1/20 ,x= 9/20