Solve 3x^2+7x=13-x,x

$$3 x ^ { 2 } + 7 x = 13 - x , x$$

Solve for x

$$x = \frac{\sqrt{55} - 4}{3} \approx 1.138732829$$

$$x=\frac{-\sqrt{55}-4}{3}\approx -3.805399496$$

Steps Using the Quadratic Formula

Subtract 13 from both sides.

$$3x^{2}+7x-13=-x$$

Add x to both sides.

$$3x^{2}+7x-13+x=0$$

Combine 7x and x to get 8x.

$$3x^{2}+8x-13=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 8 for b, and -13 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\times 3\left(-13\right)}}{2\times 3}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\times 3\left(-13\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-8±\sqrt{64-12\left(-13\right)}}{2\times 3}$$

Multiply -12 times -13.

$$x=\frac{-8±\sqrt{64+156}}{2\times 3}$$

Add 64 to 156.

$$x=\frac{-8±\sqrt{220}}{2\times 3}$$

Take the square root of 220.

$$x=\frac{-8±2\sqrt{55}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-8±2\sqrt{55}}{6}$$

Now solve the equation x=\frac{-8±2\sqrt{55}}{6} when ± is plus. Add -8 to 2\sqrt{55}.

$$x=\frac{2\sqrt{55}-8}{6}$$

Divide -8+2\sqrt{55} by 6.

$$x=\frac{\sqrt{55}-4}{3}$$

Now solve the equation x=\frac{-8±2\sqrt{55}}{6} when ± is minus. Subtract 2\sqrt{55} from -8.

$$x=\frac{-2\sqrt{55}-8}{6}$$

Divide -8-2\sqrt{55} by 6.

$$x=\frac{-\sqrt{55}-4}{3}$$

The equation is now solved.

$$x=\frac{\sqrt{55}-4}{3} x=\frac{-\sqrt{55}-4}{3}$$

Steps for Completing the Square

Add x to both sides.

$$3x^{2}+7x+x=13$$

Combine 7x and x to get 8x.

$$3x^{2}+8x=13$$

Divide both sides by 3.

$$\frac{3x^{2}+8x}{3}=\frac{13}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{8}{3}x=\frac{13}{3}$$

Divide \frac{8}{3}, the coefficient of the x term, by 2 to get \frac{4}{3}. Then add the square of \frac{4}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{8}{3}x+\left(\frac{4}{3}\right)^{2}=\frac{13}{3}+\left(\frac{4}{3}\right)^{2}$$

Square \frac{4}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{8}{3}x+\frac{16}{9}=\frac{13}{3}+\frac{16}{9}$$

Add \frac{13}{3} to \frac{16}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{8}{3}x+\frac{16}{9}=\frac{55}{9}$$

Factor x^{2}+\frac{8}{3}x+\frac{16}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{4}{3}\right)^{2}=\frac{55}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{4}{3}\right)^{2}}=\sqrt{\frac{55}{9}}$$

Simplify.

$$x+\frac{4}{3}=\frac{\sqrt{55}}{3} x+\frac{4}{3}=-\frac{\sqrt{55}}{3}$$

Subtract \frac{4}{3} from both sides of the equation.

$$x=\frac{\sqrt{55}-4}{3} x=\frac{-\sqrt{55}-4}{3}$$