Solve 320x+5120-320x-x^2-11x=0

$$320 x + 5120 - 320 x - x ^ { 2 } - 11 x = 0$$

Solve for x

$$x = \frac{3 \sqrt{2289} - 11}{2} \approx 66.265242283$$

$$x=\frac{-3\sqrt{2289}-11}{2}\approx -77.265242283$$

Steps Using the Quadratic Formula

Combine 320x and -320x to get 0.

$$5120-x^{2}-11x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-x^{2}-11x+5120=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -11 for b, and 5120 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\left(-1\right)\times 5120}}{2\left(-1\right)}$$

Square -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\left(-1\right)\times 5120}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-11\right)±\sqrt{121+4\times 5120}}{2\left(-1\right)}$$

Multiply 4 times 5120.

$$x=\frac{-\left(-11\right)±\sqrt{121+20480}}{2\left(-1\right)}$$

Add 121 to 20480.

$$x=\frac{-\left(-11\right)±\sqrt{20601}}{2\left(-1\right)}$$

Take the square root of 20601.

$$x=\frac{-\left(-11\right)±3\sqrt{2289}}{2\left(-1\right)}$$

The opposite of -11 is 11.

$$x=\frac{11±3\sqrt{2289}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{11±3\sqrt{2289}}{-2}$$

Now solve the equation x=\frac{11±3\sqrt{2289}}{-2} when ± is plus. Add 11 to 3\sqrt{2289}.

$$x=\frac{3\sqrt{2289}+11}{-2}$$

Divide 11+3\sqrt{2289} by -2.

$$x=\frac{-3\sqrt{2289}-11}{2}$$

Now solve the equation x=\frac{11±3\sqrt{2289}}{-2} when ± is minus. Subtract 3\sqrt{2289} from 11.

$$x=\frac{11-3\sqrt{2289}}{-2}$$

Divide 11-3\sqrt{2289} by -2.

$$x=\frac{3\sqrt{2289}-11}{2}$$

The equation is now solved.

$$x=\frac{-3\sqrt{2289}-11}{2} x=\frac{3\sqrt{2289}-11}{2}$$

Steps for Completing the Square

Combine 320x and -320x to get 0.

$$5120-x^{2}-11x=0$$

Subtract 5120 from both sides. Anything subtracted from zero gives its negation.

$$-x^{2}-11x=-5120$$

Divide both sides by -1.

$$\frac{-x^{2}-11x}{-1}=-\frac{5120}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{11}{-1}\right)x=-\frac{5120}{-1}$$

Divide -11 by -1.

$$x^{2}+11x=-\frac{5120}{-1}$$

Divide -5120 by -1.

$$x^{2}+11x=5120$$

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.

$$x^{2}+11x+\left(\frac{11}{2}\right)^{2}=5120+\left(\frac{11}{2}\right)^{2}$$

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

$$x^{2}+11x+\frac{121}{4}=5120+\frac{121}{4}$$

Add 5120 to \frac{121}{4}.

$$x^{2}+11x+\frac{121}{4}=\frac{20601}{4}$$

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}.

$$\left(x+\frac{11}{2}\right)^{2}=\frac{20601}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{11}{2}\right)^{2}}=\sqrt{\frac{20601}{4}}$$

Simplify.

$$x+\frac{11}{2}=\frac{3\sqrt{2289}}{2} x+\frac{11}{2}=-\frac{3\sqrt{2289}}{2}$$

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

$$x=\frac{3\sqrt{2289}-11}{2} x=\frac{-3\sqrt{2289}-11}{2}$$