Solve 3x^2=-x+5

$$3 x ^ { 2 } = - x + 5$$

Solve for x

$$x = \frac{\sqrt{61} - 1}{6} \approx 1.135041613$$

$$x=\frac{-\sqrt{61}-1}{6}\approx -1.468374946$$

Steps Using the Quadratic Formula

Add x to both sides.

$$3x^{2}+x=5$$

Subtract 5 from both sides.

$$3x^{2}+x-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 1 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\times 3\left(-5\right)}}{2\times 3}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\times 3\left(-5\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-1±\sqrt{1-12\left(-5\right)}}{2\times 3}$$

Multiply -12 times -5.

$$x=\frac{-1±\sqrt{1+60}}{2\times 3}$$

Add 1 to 60.

$$x=\frac{-1±\sqrt{61}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-1±\sqrt{61}}{6}$$

Now solve the equation x=\frac{-1±\sqrt{61}}{6} when ± is plus. Add -1 to \sqrt{61}.

$$x=\frac{\sqrt{61}-1}{6}$$

Now solve the equation x=\frac{-1±\sqrt{61}}{6} when ± is minus. Subtract \sqrt{61} from -1.

$$x=\frac{-\sqrt{61}-1}{6}$$

The equation is now solved.

$$x=\frac{\sqrt{61}-1}{6} x=\frac{-\sqrt{61}-1}{6}$$

Steps for Completing the Square

Add x to both sides.

$$3x^{2}+x=5$$

Divide both sides by 3.

$$\frac{3x^{2}+x}{3}=\frac{5}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{1}{3}x=\frac{5}{3}$$

Divide \frac{1}{3}, the coefficient of the x term, by 2 to get \frac{1}{6}. Then add the square of \frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{3}x+\left(\frac{1}{6}\right)^{2}=\frac{5}{3}+\left(\frac{1}{6}\right)^{2}$$

Square \frac{1}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{3}x+\frac{1}{36}=\frac{5}{3}+\frac{1}{36}$$

Add \frac{5}{3} to \frac{1}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{3}x+\frac{1}{36}=\frac{61}{36}$$

Factor x^{2}+\frac{1}{3}x+\frac{1}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{6}\right)^{2}=\frac{61}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{6}\right)^{2}}=\sqrt{\frac{61}{36}}$$

Simplify.

$$x+\frac{1}{6}=\frac{\sqrt{61}}{6} x+\frac{1}{6}=-\frac{\sqrt{61}}{6}$$

Subtract \frac{1}{6} from both sides of the equation.

$$x=\frac{\sqrt{61}-1}{6} x=\frac{-\sqrt{61}-1}{6}$$