Solve 264=3x+x^2

$$264 = 3 x + x ^ { 2 }$$

Solve for x

$$x = \frac{\sqrt{1065} - 3}{2} \approx 14.817168872$$

$$x=\frac{-\sqrt{1065}-3}{2}\approx -17.817168872$$

Steps Using the Quadratic Formula

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

$$3x+x^{2}=264$$

Subtract 264 from both sides.

$$3x+x^{2}-264=0$$

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.

$$x^{2}+3x-264=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 3 for b, and -264 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-264\right)}}{2}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-264\right)}}{2}$$

Multiply -4 times -264.

$$x=\frac{-3±\sqrt{9+1056}}{2}$$

Add 9 to 1056.

$$x=\frac{-3±\sqrt{1065}}{2}$$

Now solve the equation x=\frac{-3±\sqrt{1065}}{2} when ± is plus. Add -3 to \sqrt{1065}.

$$x=\frac{\sqrt{1065}-3}{2}$$

Now solve the equation x=\frac{-3±\sqrt{1065}}{2} when ± is minus. Subtract \sqrt{1065} from -3.

$$x=\frac{-\sqrt{1065}-3}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{1065}-3}{2} x=\frac{-\sqrt{1065}-3}{2}$$

Steps for Completing the Square

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

$$3x+x^{2}=264$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$x^{2}+3x=264$$

Divide 3, the coefficient of the x term, by 2 to get \frac{3}{2}. Then add the square of \frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=264+\left(\frac{3}{2}\right)^{2}$$

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+3x+\frac{9}{4}=264+\frac{9}{4}$$

Add 264 to \frac{9}{4}.

$$x^{2}+3x+\frac{9}{4}=\frac{1065}{4}$$

Factor x^{2}+3x+\frac{9}{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}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{1065}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{1065}{4}}$$

Simplify.

$$x+\frac{3}{2}=\frac{\sqrt{1065}}{2} x+\frac{3}{2}=-\frac{\sqrt{1065}}{2}$$

Subtract \frac{3}{2} from both sides of the equation.

$$x=\frac{\sqrt{1065}-3}{2} x=\frac{-\sqrt{1065}-3}{2}$$