Solve x^2+{left(-2-3x/4right)}^2-2x=0

$${ x }^{ 2 } + { \left( \frac{ -2-3x }{ 4 } \right) }^{ 2 } -2x=0$$

Solve for x

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

Steps Using the Quadratic Formula

To raise \frac{-2-3x}{4} to a power, raise both numerator and denominator to the power and then divide.

$$x^{2}+\frac{\left(-2-3x\right)^{2}}{4^{2}}-2x=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x^{2}-2x times \frac{4^{2}}{4^{2}}.

$$\frac{\left(x^{2}-2x\right)\times 4^{2}}{4^{2}}+\frac{\left(-2-3x\right)^{2}}{4^{2}}=0$$

Since \frac{\left(x^{2}-2x\right)\times 4^{2}}{4^{2}} and \frac{\left(-2-3x\right)^{2}}{4^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(x^{2}-2x\right)\times 4^{2}+\left(-2-3x\right)^{2}}{4^{2}}=0$$

Do the multiplications in \left(x^{2}-2x\right)\times 4^{2}+\left(-2-3x\right)^{2}.

$$\frac{16x^{2}-32x+4+12x+9x^{2}}{4^{2}}=0$$

Combine like terms in 16x^{2}-32x+4+12x+9x^{2}.

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

Calculate 4 to the power of 2 and get 16.

$$\frac{25x^{2}-20x+4}{16}=0$$

Divide each term of 25x^{2}-20x+4 by 16 to get \frac{25}{16}x^{2}-\frac{5}{4}x+\frac{1}{4}.

$$\frac{25}{16}x^{2}-\frac{5}{4}x+\frac{1}{4}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{25}{16} for a, -\frac{5}{4} for b, and \frac{1}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{5}{4}\right)±\sqrt{\left(-\frac{5}{4}\right)^{2}-4\times \frac{25}{16}\times \frac{1}{4}}}{2\times \frac{25}{16}}$$

Square -\frac{5}{4} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-\frac{5}{4}\right)±\sqrt{\frac{25}{16}-4\times \frac{25}{16}\times \frac{1}{4}}}{2\times \frac{25}{16}}$$

Multiply -4 times \frac{25}{16}.

$$x=\frac{-\left(-\frac{5}{4}\right)±\sqrt{\frac{25}{16}-\frac{25}{4}\times \frac{1}{4}}}{2\times \frac{25}{16}}$$

Multiply -\frac{25}{4} times \frac{1}{4} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{5}{4}\right)±\sqrt{\frac{25-25}{16}}}{2\times \frac{25}{16}}$$

Add \frac{25}{16} to -\frac{25}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{5}{4}\right)±\sqrt{0}}{2\times \frac{25}{16}}$$

Take the square root of 0.

$$x=-\frac{-\frac{5}{4}}{2\times \frac{25}{16}}$$

The opposite of -\frac{5}{4} is \frac{5}{4}.

$$x=\frac{\frac{5}{4}}{2\times \frac{25}{16}}$$

Multiply 2 times \frac{25}{16}.

$$x=\frac{\frac{5}{4}}{\frac{25}{8}}$$

Divide \frac{5}{4} by \frac{25}{8} by multiplying \frac{5}{4} by the reciprocal of \frac{25}{8}.

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

Steps for Completing the Square

To raise \frac{-2-3x}{4} to a power, raise both numerator and denominator to the power and then divide.

$$x^{2}+\frac{\left(-2-3x\right)^{2}}{4^{2}}-2x=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x^{2}-2x times \frac{4^{2}}{4^{2}}.

$$\frac{\left(x^{2}-2x\right)\times 4^{2}}{4^{2}}+\frac{\left(-2-3x\right)^{2}}{4^{2}}=0$$

Since \frac{\left(x^{2}-2x\right)\times 4^{2}}{4^{2}} and \frac{\left(-2-3x\right)^{2}}{4^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(x^{2}-2x\right)\times 4^{2}+\left(-2-3x\right)^{2}}{4^{2}}=0$$

Do the multiplications in \left(x^{2}-2x\right)\times 4^{2}+\left(-2-3x\right)^{2}.

$$\frac{16x^{2}-32x+4+12x+9x^{2}}{4^{2}}=0$$

Combine like terms in 16x^{2}-32x+4+12x+9x^{2}.

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

Calculate 4 to the power of 2 and get 16.

$$\frac{25x^{2}-20x+4}{16}=0$$

Divide each term of 25x^{2}-20x+4 by 16 to get \frac{25}{16}x^{2}-\frac{5}{4}x+\frac{1}{4}.

$$\frac{25}{16}x^{2}-\frac{5}{4}x+\frac{1}{4}=0$$

Subtract \frac{1}{4} from both sides. Anything subtracted from zero gives its negation.

$$\frac{25}{16}x^{2}-\frac{5}{4}x=-\frac{1}{4}$$

Divide both sides of the equation by \frac{25}{16}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{25}{16}x^{2}-\frac{5}{4}x}{\frac{25}{16}}=-\frac{\frac{1}{4}}{\frac{25}{16}}$$

Dividing by \frac{25}{16} undoes the multiplication by \frac{25}{16}.

$$x^{2}+\left(-\frac{\frac{5}{4}}{\frac{25}{16}}\right)x=-\frac{\frac{1}{4}}{\frac{25}{16}}$$

Divide -\frac{5}{4} by \frac{25}{16} by multiplying -\frac{5}{4} by the reciprocal of \frac{25}{16}.

$$x^{2}-\frac{4}{5}x=-\frac{\frac{1}{4}}{\frac{25}{16}}$$

Divide -\frac{1}{4} by \frac{25}{16} by multiplying -\frac{1}{4} by the reciprocal of \frac{25}{16}.

$$x^{2}-\frac{4}{5}x=-\frac{4}{25}$$

Divide -\frac{4}{5}, the coefficient of the x term, by 2 to get -\frac{2}{5}. Then add the square of -\frac{2}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{4}{5}x+\left(-\frac{2}{5}\right)^{2}=-\frac{4}{25}+\left(-\frac{2}{5}\right)^{2}$$

Square -\frac{2}{5} by squaring both the numerator and the denominator of the fraction.

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

Add -\frac{4}{25} to \frac{4}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{4}{5}x+\frac{4}{25}=0$$

Factor x^{2}-\frac{4}{5}x+\frac{4}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{2}{5}\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{5}\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-\frac{2}{5}=0 x-\frac{2}{5}=0$$

Add \frac{2}{5} to both sides of the equation.

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

The equation is now solved. Solutions are the same.

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