Solve (4x+2)^2=4x(4x+8)

$$( 4 x + 2 ) ^ { 2 } = 4 x ( 4 x + 8 )$$

Solve for x

$$x=\frac{1}{4}=0.25$$

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4x+2\right)^{2}.

$$16x^{2}+16x+4=4x\left(4x+8\right)$$

Use the distributive property to multiply 4x by 4x+8.

$$16x^{2}+16x+4=16x^{2}+32x$$

Subtract 16x^{2} from both sides.

$$16x^{2}+16x+4-16x^{2}=32x$$

Combine 16x^{2} and -16x^{2} to get 0.

$$16x+4=32x$$

Subtract 32x from both sides.

$$16x+4-32x=0$$

Combine 16x and -32x to get -16x.

$$-16x+4=0$$

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

$$-16x=-4$$

Divide both sides by -16.

$$x=\frac{-4}{-16}$$

Reduce the fraction \frac{-4}{-16} to lowest terms by extracting and canceling out -4.

$$x=\frac{1}{4}$$