Solve 0=11x^2+121x+308

$$0 = 11 x ^ { 2 } + 121 x + 308$$

Solve for x

$$x=-7$$

$$x=-4$$

Steps Using the Quadratic Formula

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

$$11x^{2}+121x+308=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 11 for a, 121 for b, and 308 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-121±\sqrt{121^{2}-4\times 11\times 308}}{2\times 11}$$

Square 121.

$$x=\frac{-121±\sqrt{14641-4\times 11\times 308}}{2\times 11}$$

Multiply -4 times 11.

$$x=\frac{-121±\sqrt{14641-44\times 308}}{2\times 11}$$

Multiply -44 times 308.

$$x=\frac{-121±\sqrt{14641-13552}}{2\times 11}$$

Add 14641 to -13552.

$$x=\frac{-121±\sqrt{1089}}{2\times 11}$$

Take the square root of 1089.

$$x=\frac{-121±33}{2\times 11}$$

Multiply 2 times 11.

$$x=\frac{-121±33}{22}$$

Now solve the equation x=\frac{-121±33}{22} when ± is plus. Add -121 to 33.

$$x=-\frac{88}{22}$$

Divide -88 by 22.

$$x=-4$$

Now solve the equation x=\frac{-121±33}{22} when ± is minus. Subtract 33 from -121.

$$x=-\frac{154}{22}$$

Divide -154 by 22.

$$x=-7$$

The equation is now solved.

$$x=-4 x=-7$$

Steps for Completing the Square

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

$$11x^{2}+121x+308=0$$

Subtract 308 from both sides. Anything subtracted from zero gives its negation.

$$11x^{2}+121x=-308$$

Divide both sides by 11.

$$\frac{11x^{2}+121x}{11}=-\frac{308}{11}$$

Dividing by 11 undoes the multiplication by 11.

$$x^{2}+\frac{121}{11}x=-\frac{308}{11}$$

Divide 121 by 11.

$$x^{2}+11x=-\frac{308}{11}$$

Divide -308 by 11.

$$x^{2}+11x=-28$$

Divide 11, the coefficient of the x term, by 2 to get \frac{11}{2}. Then add the square of \frac{11}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+11x+\left(\frac{11}{2}\right)^{2}=-28+\left(\frac{11}{2}\right)^{2}$$

Square \frac{11}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+11x+\frac{121}{4}=-28+\frac{121}{4}$$

Add -28 to \frac{121}{4}.

$$x^{2}+11x+\frac{121}{4}=\frac{9}{4}$$

Factor x^{2}+11x+\frac{121}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{11}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{11}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

$$x+\frac{11}{2}=\frac{3}{2} x+\frac{11}{2}=-\frac{3}{2}$$

Subtract \frac{11}{2} from both sides of the equation.

$$x=-4 x=-7$$