What are the solutions to the equation (4x+1)^2-8=1 ?

The solutions to the equation (4x+1)^2-8=1 are x= 1/2 ,x=-1

Find the zeros of (4x+1)^2-8=1

The zeros of (4x+1)^2-8=1 are x= 1/2 ,x=-1

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(4 x + 1)^{2} - 8 = 1

x = \frac{1}{2}, x = - 1

Decimal Notation

x = 0.5, x = - 1

Solution steps

(4 x + 1)^{2} - 8 = 1

Move

8

to the right side

(4 x + 1)^{2} = 9

For

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

the solutions are

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

Solve

4 x + 1 = 9 : x = \frac{1}{2}

Solve

4 x + 1 = - 9 : x = - 1

The solutions to the quadratic equation are:

x = \frac{1}{2}, x = - 1

x, x^{2} - 4 x + 5 = 0

v, \frac{1}{2} m v^{2} = m g h

x^{2} + 0.13 x + 0.0032 = 0

y, 2 x^{3} + x^{2} y + y^{2} = 4

0 = 3 x^{2} - 12 x + 12

What are the solutions to the equation (4x+1)^2-8=1 ?
The solutions to the equation (4x+1)^2-8=1 are x= 1/2 ,x=-1
Find the zeros of (4x+1)^2-8=1
The zeros of (4x+1)^2-8=1 are x= 1/2 ,x=-1