Solve 4=-1/5x^2-21/10x

$$4 = - \frac { 1 } { 5 } x ^ { 2 } - \frac { 21 } { 10 } x$$

Solve for x

$$x=-8$$

$$x = -\frac{5}{2} = -2\frac{1}{2} = -2.5$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$-\frac{1}{5}x^{2}-\frac{21}{10}x=4$$

Subtract 4 from both sides.

$$-\frac{1}{5}x^{2}-\frac{21}{10}x-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{1}{5} for a, -\frac{21}{10} for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{21}{10}\right)±\sqrt{\left(-\frac{21}{10}\right)^{2}-4\left(-\frac{1}{5}\right)\left(-4\right)}}{2\left(-\frac{1}{5}\right)}$$

Square -\frac{21}{10} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-\frac{21}{10}\right)±\sqrt{\frac{441}{100}-4\left(-\frac{1}{5}\right)\left(-4\right)}}{2\left(-\frac{1}{5}\right)}$$

Multiply -4 times -\frac{1}{5}.

$$x=\frac{-\left(-\frac{21}{10}\right)±\sqrt{\frac{441}{100}+\frac{4}{5}\left(-4\right)}}{2\left(-\frac{1}{5}\right)}$$

Multiply \frac{4}{5} times -4.

$$x=\frac{-\left(-\frac{21}{10}\right)±\sqrt{\frac{441}{100}-\frac{16}{5}}}{2\left(-\frac{1}{5}\right)}$$

Add \frac{441}{100} to -\frac{16}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{21}{10}\right)±\sqrt{\frac{121}{100}}}{2\left(-\frac{1}{5}\right)}$$

Take the square root of \frac{121}{100}.

$$x=\frac{-\left(-\frac{21}{10}\right)±\frac{11}{10}}{2\left(-\frac{1}{5}\right)}$$

The opposite of -\frac{21}{10} is \frac{21}{10}.

$$x=\frac{\frac{21}{10}±\frac{11}{10}}{2\left(-\frac{1}{5}\right)}$$

Multiply 2 times -\frac{1}{5}.

$$x=\frac{\frac{21}{10}±\frac{11}{10}}{-\frac{2}{5}}$$

Now solve the equation x=\frac{\frac{21}{10}±\frac{11}{10}}{-\frac{2}{5}} when ± is plus. Add \frac{21}{10} to \frac{11}{10} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{\frac{16}{5}}{-\frac{2}{5}}$$

Divide \frac{16}{5} by -\frac{2}{5} by multiplying \frac{16}{5} by the reciprocal of -\frac{2}{5}.

$$x=-8$$

Now solve the equation x=\frac{\frac{21}{10}±\frac{11}{10}}{-\frac{2}{5}} when ± is minus. Subtract \frac{11}{10} from \frac{21}{10} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{1}{-\frac{2}{5}}$$

Divide 1 by -\frac{2}{5} by multiplying 1 by the reciprocal of -\frac{2}{5}.

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

The equation is now solved.

$$x=-8 x=-\frac{5}{2}$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$-\frac{1}{5}x^{2}-\frac{21}{10}x=4$$

Multiply both sides by -5.

$$\frac{-\frac{1}{5}x^{2}-\frac{21}{10}x}{-\frac{1}{5}}=\frac{4}{-\frac{1}{5}}$$

Dividing by -\frac{1}{5} undoes the multiplication by -\frac{1}{5}.

$$x^{2}+\left(-\frac{\frac{21}{10}}{-\frac{1}{5}}\right)x=\frac{4}{-\frac{1}{5}}$$

Divide -\frac{21}{10} by -\frac{1}{5} by multiplying -\frac{21}{10} by the reciprocal of -\frac{1}{5}.

$$x^{2}+\frac{21}{2}x=\frac{4}{-\frac{1}{5}}$$

Divide 4 by -\frac{1}{5} by multiplying 4 by the reciprocal of -\frac{1}{5}.

$$x^{2}+\frac{21}{2}x=-20$$

Divide \frac{21}{2}, the coefficient of the x term, by 2 to get \frac{21}{4}. Then add the square of \frac{21}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{21}{2}x+\left(\frac{21}{4}\right)^{2}=-20+\left(\frac{21}{4}\right)^{2}$$

Square \frac{21}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{21}{2}x+\frac{441}{16}=-20+\frac{441}{16}$$

Add -20 to \frac{441}{16}.

$$x^{2}+\frac{21}{2}x+\frac{441}{16}=\frac{121}{16}$$

Factor x^{2}+\frac{21}{2}x+\frac{441}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{21}{4}\right)^{2}=\frac{121}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{21}{4}\right)^{2}}=\sqrt{\frac{121}{16}}$$

Simplify.

$$x+\frac{21}{4}=\frac{11}{4} x+\frac{21}{4}=-\frac{11}{4}$$

Subtract \frac{21}{4} from both sides of the equation.

$$x=-\frac{5}{2} x=-8$$