Solve -5=-4q^2-5q

$$- 5 = - 4 q ^ { 2 } - 5 q$$

Solve for q

$$q=\frac{\sqrt{105}-5}{8}\approx 0.655868846$$

$$q=\frac{-\sqrt{105}-5}{8}\approx -1.905868846$$

Steps Using the Quadratic Formula

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

$$-4q^{2}-5q=-5$$

Add 5 to both sides.

$$-4q^{2}-5q+5=0$$

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

$$q=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-4\right)\times 5}}{2\left(-4\right)}$$

Square -5.

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

Multiply -4 times -4.

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

Multiply 16 times 5.

$$q=\frac{-\left(-5\right)±\sqrt{25+80}}{2\left(-4\right)}$$

Add 25 to 80.

$$q=\frac{-\left(-5\right)±\sqrt{105}}{2\left(-4\right)}$$

The opposite of -5 is 5.

$$q=\frac{5±\sqrt{105}}{2\left(-4\right)}$$

Multiply 2 times -4.

$$q=\frac{5±\sqrt{105}}{-8}$$

Now solve the equation q=\frac{5±\sqrt{105}}{-8} when ± is plus. Add 5 to \sqrt{105}.

$$q=\frac{\sqrt{105}+5}{-8}$$

Divide 5+\sqrt{105} by -8.

$$q=\frac{-\sqrt{105}-5}{8}$$

Now solve the equation q=\frac{5±\sqrt{105}}{-8} when ± is minus. Subtract \sqrt{105} from 5.

$$q=\frac{5-\sqrt{105}}{-8}$$

Divide 5-\sqrt{105} by -8.

$$q=\frac{\sqrt{105}-5}{8}$$

The equation is now solved.

$$q=\frac{-\sqrt{105}-5}{8} q=\frac{\sqrt{105}-5}{8}$$

Steps for Completing the Square

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

$$-4q^{2}-5q=-5$$

Divide both sides by -4.

$$\frac{-4q^{2}-5q}{-4}=-\frac{5}{-4}$$

Dividing by -4 undoes the multiplication by -4.

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

Divide -5 by -4.

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

Divide -5 by -4.

$$q^{2}+\frac{5}{4}q=\frac{5}{4}$$

Divide \frac{5}{4}, the coefficient of the x term, by 2 to get \frac{5}{8}. Then add the square of \frac{5}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$q^{2}+\frac{5}{4}q+\left(\frac{5}{8}\right)^{2}=\frac{5}{4}+\left(\frac{5}{8}\right)^{2}$$

Square \frac{5}{8} by squaring both the numerator and the denominator of the fraction.

$$q^{2}+\frac{5}{4}q+\frac{25}{64}=\frac{5}{4}+\frac{25}{64}$$

Add \frac{5}{4} to \frac{25}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$q^{2}+\frac{5}{4}q+\frac{25}{64}=\frac{105}{64}$$

Factor q^{2}+\frac{5}{4}q+\frac{25}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(q+\frac{5}{8}\right)^{2}=\frac{105}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(q+\frac{5}{8}\right)^{2}}=\sqrt{\frac{105}{64}}$$

Simplify.

$$q+\frac{5}{8}=\frac{\sqrt{105}}{8} q+\frac{5}{8}=-\frac{\sqrt{105}}{8}$$

Subtract \frac{5}{8} from both sides of the equation.

$$q=\frac{\sqrt{105}-5}{8} q=\frac{-\sqrt{105}-5}{8}$$