Solve 1+3x^2+6x-9=0

$$1+3 { x }^{ 2 } +6x-9 = 0$$

Solve for x

$$x=\frac{\sqrt{33}}{3}-1\approx 0.914854216$$

$$x=-\frac{\sqrt{33}}{3}-1\approx -2.914854216$$

Steps Using the Quadratic Formula

Subtract 9 from 1 to get -8.

$$-8+3x^{2}+6x=0$$

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.

$$3x^{2}+6x-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 6 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\times 3\left(-8\right)}}{2\times 3}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\times 3\left(-8\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-6±\sqrt{36-12\left(-8\right)}}{2\times 3}$$

Multiply -12 times -8.

$$x=\frac{-6±\sqrt{36+96}}{2\times 3}$$

Add 36 to 96.

$$x=\frac{-6±\sqrt{132}}{2\times 3}$$

Take the square root of 132.

$$x=\frac{-6±2\sqrt{33}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-6±2\sqrt{33}}{6}$$

Now solve the equation x=\frac{-6±2\sqrt{33}}{6} when ± is plus. Add -6 to 2\sqrt{33}.

$$x=\frac{2\sqrt{33}-6}{6}$$

Divide -6+2\sqrt{33} by 6.

$$x=\frac{\sqrt{33}}{3}-1$$

Now solve the equation x=\frac{-6±2\sqrt{33}}{6} when ± is minus. Subtract 2\sqrt{33} from -6.

$$x=\frac{-2\sqrt{33}-6}{6}$$

Divide -6-2\sqrt{33} by 6.

$$x=-\frac{\sqrt{33}}{3}-1$$

The equation is now solved.

$$x=\frac{\sqrt{33}}{3}-1 x=-\frac{\sqrt{33}}{3}-1$$

Steps for Completing the Square

Subtract 9 from 1 to get -8.

$$-8+3x^{2}+6x=0$$

Add 8 to both sides. Anything plus zero gives itself.

$$3x^{2}+6x=8$$

Divide both sides by 3.

$$\frac{3x^{2}+6x}{3}=\frac{8}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{6}{3}x=\frac{8}{3}$$

Divide 6 by 3.

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

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

$$x^{2}+2x+1^{2}=\frac{8}{3}+1^{2}$$

Square 1.

$$x^{2}+2x+1=\frac{8}{3}+1$$

Add \frac{8}{3} to 1.

$$x^{2}+2x+1=\frac{11}{3}$$

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

$$\left(x+1\right)^{2}=\frac{11}{3}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{\frac{11}{3}}$$

Simplify.

$$x+1=\frac{\sqrt{33}}{3} x+1=-\frac{\sqrt{33}}{3}$$

Subtract 1 from both sides of the equation.

$$x=\frac{\sqrt{33}}{3}-1 x=-\frac{\sqrt{33}}{3}-1$$