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

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

Solve for x (complex solution)

$$x=\frac{2}{5}i=0.4i$$

$$x=-\frac{2}{5}i=-0.4i$$

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}.

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

Add 20x to both sides.

$$25x^{2}-20x+4+20x=0$$

Combine -20x and 20x to get 0.

$$25x^{2}+4=0$$

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

$$25x^{2}=-4$$

Divide both sides by 25.

$$x^{2}=-\frac{4}{25}$$

The equation is now solved.

$$x=\frac{2}{5}i x=-\frac{2}{5}i$$

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}.

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

Add 20x to both sides.

$$25x^{2}-20x+4+20x=0$$

Combine -20x and 20x to get 0.

$$25x^{2}+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 25 for a, 0 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 25\times 4}}{2\times 25}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 25\times 4}}{2\times 25}$$

Multiply -4 times 25.

$$x=\frac{0±\sqrt{-100\times 4}}{2\times 25}$$

Multiply -100 times 4.

$$x=\frac{0±\sqrt{-400}}{2\times 25}$$

Take the square root of -400.

$$x=\frac{0±20i}{2\times 25}$$

Multiply 2 times 25.

$$x=\frac{0±20i}{50}$$

Now solve the equation x=\frac{0±20i}{50} when ± is plus.

$$x=\frac{2}{5}i$$

Now solve the equation x=\frac{0±20i}{50} when ± is minus.

$$x=-\frac{2}{5}i$$

The equation is now solved.

$$x=\frac{2}{5}i x=-\frac{2}{5}i$$