Solve (c)(4c+15)+4=0

$$( c ) ( 4 c + 15 ) + 4 = 0$$

Solve for c

$$c=\frac{\sqrt{161}-15}{8}\approx -0.288927807$$

$$c=\frac{-\sqrt{161}-15}{8}\approx -3.461072193$$

Steps Using the Quadratic Formula

Use the distributive property to multiply c by 4c+15.

$$4c^{2}+15c+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 15 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$c=\frac{-15±\sqrt{15^{2}-4\times 4\times 4}}{2\times 4}$$

Square 15.

$$c=\frac{-15±\sqrt{225-4\times 4\times 4}}{2\times 4}$$

Multiply -4 times 4.

$$c=\frac{-15±\sqrt{225-16\times 4}}{2\times 4}$$

Multiply -16 times 4.

$$c=\frac{-15±\sqrt{225-64}}{2\times 4}$$

Add 225 to -64.

$$c=\frac{-15±\sqrt{161}}{2\times 4}$$

Multiply 2 times 4.

$$c=\frac{-15±\sqrt{161}}{8}$$

Now solve the equation c=\frac{-15±\sqrt{161}}{8} when ± is plus. Add -15 to \sqrt{161}.

$$c=\frac{\sqrt{161}-15}{8}$$

Now solve the equation c=\frac{-15±\sqrt{161}}{8} when ± is minus. Subtract \sqrt{161} from -15.

$$c=\frac{-\sqrt{161}-15}{8}$$

The equation is now solved.

$$c=\frac{\sqrt{161}-15}{8} c=\frac{-\sqrt{161}-15}{8}$$

Steps for Completing the Square

Use the distributive property to multiply c by 4c+15.

$$4c^{2}+15c+4=0$$

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

$$4c^{2}+15c=-4$$

Divide both sides by 4.

$$\frac{4c^{2}+15c}{4}=-\frac{4}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$c^{2}+\frac{15}{4}c=-\frac{4}{4}$$

Divide -4 by 4.

$$c^{2}+\frac{15}{4}c=-1$$

Divide \frac{15}{4}, the coefficient of the x term, by 2 to get \frac{15}{8}. Then add the square of \frac{15}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$c^{2}+\frac{15}{4}c+\left(\frac{15}{8}\right)^{2}=-1+\left(\frac{15}{8}\right)^{2}$$

Square \frac{15}{8} by squaring both the numerator and the denominator of the fraction.

$$c^{2}+\frac{15}{4}c+\frac{225}{64}=-1+\frac{225}{64}$$

Add -1 to \frac{225}{64}.

$$c^{2}+\frac{15}{4}c+\frac{225}{64}=\frac{161}{64}$$

Factor c^{2}+\frac{15}{4}c+\frac{225}{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}.

$$\left(c+\frac{15}{8}\right)^{2}=\frac{161}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(c+\frac{15}{8}\right)^{2}}=\sqrt{\frac{161}{64}}$$

Simplify.

$$c+\frac{15}{8}=\frac{\sqrt{161}}{8} c+\frac{15}{8}=-\frac{\sqrt{161}}{8}$$

Subtract \frac{15}{8} from both sides of the equation.

$$c=\frac{\sqrt{161}-15}{8} c=\frac{-\sqrt{161}-15}{8}$$