Solve 3x^2=-9x+2

$$3 x ^ { 2 } = - 9 x + 2$$

Solve for x

$$x=\frac{\sqrt{105}}{6}-\frac{3}{2}\approx 0.207825128$$

$$x=-\frac{\sqrt{105}}{6}-\frac{3}{2}\approx -3.207825128$$

Steps Using the Quadratic Formula

Add 9x to both sides.

$$3x^{2}+9x=2$$

Subtract 2 from both sides.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 9 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 9.

$$x=\frac{-9±\sqrt{81-4\times 3\left(-2\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-9±\sqrt{81-12\left(-2\right)}}{2\times 3}$$

Multiply -12 times -2.

$$x=\frac{-9±\sqrt{81+24}}{2\times 3}$$

Add 81 to 24.

$$x=\frac{-9±\sqrt{105}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-9±\sqrt{105}}{6}$$

Now solve the equation x=\frac{-9±\sqrt{105}}{6} when ± is plus. Add -9 to \sqrt{105}.

$$x=\frac{\sqrt{105}-9}{6}$$

Divide -9+\sqrt{105} by 6.

$$x=\frac{\sqrt{105}}{6}-\frac{3}{2}$$

Now solve the equation x=\frac{-9±\sqrt{105}}{6} when ± is minus. Subtract \sqrt{105} from -9.

$$x=\frac{-\sqrt{105}-9}{6}$$

Divide -9-\sqrt{105} by 6.

$$x=-\frac{\sqrt{105}}{6}-\frac{3}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{105}}{6}-\frac{3}{2} x=-\frac{\sqrt{105}}{6}-\frac{3}{2}$$

Steps for Completing the Square

Add 9x to both sides.

$$3x^{2}+9x=2$$

Divide both sides by 3.

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

Dividing by 3 undoes the multiplication by 3.

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

Divide 9 by 3.

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

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.

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

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

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

Add \frac{2}{3} to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+3x+\frac{9}{4}=\frac{35}{12}$$

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}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{35}{12}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{35}{12}}$$

Simplify.

$$x+\frac{3}{2}=\frac{\sqrt{105}}{6} x+\frac{3}{2}=-\frac{\sqrt{105}}{6}$$

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

$$x=\frac{\sqrt{105}}{6}-\frac{3}{2} x=-\frac{\sqrt{105}}{6}-\frac{3}{2}$$