Solve 0=3x^2+x-3

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

Solve for x

$$x=\frac{\sqrt{37}-1}{6}\approx 0.847127088$$

$$x=\frac{-\sqrt{37}-1}{6}\approx -1.180460422$$

Steps Using the Quadratic Formula

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

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

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

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

Square 1.

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

Multiply -4 times 3.

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

Multiply -12 times -3.

$$x=\frac{-1±\sqrt{1+36}}{2\times 3}$$

Add 1 to 36.

$$x=\frac{-1±\sqrt{37}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-1±\sqrt{37}}{6}$$

Now solve the equation x=\frac{-1±\sqrt{37}}{6} when ± is plus. Add -1 to \sqrt{37}.

$$x=\frac{\sqrt{37}-1}{6}$$

Now solve the equation x=\frac{-1±\sqrt{37}}{6} when ± is minus. Subtract \sqrt{37} from -1.

$$x=\frac{-\sqrt{37}-1}{6}$$

The equation is now solved.

$$x=\frac{\sqrt{37}-1}{6} x=\frac{-\sqrt{37}-1}{6}$$

Steps for Completing the Square

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

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

Add 3 to both sides. Anything plus zero gives itself.

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

Divide both sides by 3.

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

Dividing by 3 undoes the multiplication by 3.

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

Divide 3 by 3.

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

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.

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

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

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

Add 1 to \frac{1}{36}.

$$x^{2}+\frac{1}{3}x+\frac{1}{36}=\frac{37}{36}$$

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}.

$$\left(x+\frac{1}{6}\right)^{2}=\frac{37}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{6}\right)^{2}}=\sqrt{\frac{37}{36}}$$

Simplify.

$$x+\frac{1}{6}=\frac{\sqrt{37}}{6} x+\frac{1}{6}=-\frac{\sqrt{37}}{6}$$

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

$$x=\frac{\sqrt{37}-1}{6} x=\frac{-\sqrt{37}-1}{6}$$