Solve (5x-2)^2=5x^2-20x+4

$$( 5 x - 2 ) ^ { 2 } = 5 x ^ { 2 } - 20 x + 4$$

Solve for x

$$x=0$$

Steps by Finding Square Root

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5x-2\right)^{2}.

$$25x^{2}-20x+4=5x^{2}-20x+4$$

Subtract 5x^{2} from both sides.

$$25x^{2}-20x+4-5x^{2}=-20x+4$$

Combine 25x^{2} and -5x^{2} to get 20x^{2}.

$$20x^{2}-20x+4=-20x+4$$

Add 20x to both sides.

$$20x^{2}-20x+4+20x=4$$

Combine -20x and 20x to get 0.

$$20x^{2}+4=4$$

Subtract 4 from both sides.

$$20x^{2}=4-4$$

Subtract 4 from 4 to get 0.

$$20x^{2}=0$$

Divide both sides by 20. Zero divided by any non-zero number gives zero.

$$x^{2}=0$$

Take the square root of both sides of the equation.

$$x=0 x=0$$

The equation is now solved. Solutions are the same.

$$x=0$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5x-2\right)^{2}.

$$25x^{2}-20x+4=5x^{2}-20x+4$$

Subtract 5x^{2} from both sides.

$$25x^{2}-20x+4-5x^{2}=-20x+4$$

Combine 25x^{2} and -5x^{2} to get 20x^{2}.

$$20x^{2}-20x+4=-20x+4$$

Add 20x to both sides.

$$20x^{2}-20x+4+20x=4$$

Combine -20x and 20x to get 0.

$$20x^{2}+4=4$$

Subtract 4 from both sides.

$$20x^{2}+4-4=0$$

Subtract 4 from 4 to get 0.

$$20x^{2}=0$$

Divide both sides by 20. Zero divided by any non-zero number gives zero.

$$x^{2}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}}}{2}$$

Take the square root of 0^{2}.

$$x=\frac{0±0}{2}$$

Divide 0 by 2.

$$x=0$$