Solve 3x^2=9x-5

$$3 x ^ { 2 } = 9 x - 5$$

Solve for x

$$x=\frac{\sqrt{21}}{6}+\frac{3}{2}\approx 2.263762616$$

$$x=-\frac{\sqrt{21}}{6}+\frac{3}{2}\approx 0.736237384$$

Steps Using the Quadratic Formula

Subtract 9x from both sides.

$$3x^{2}-9x=-5$$

Add 5 to both sides.

$$3x^{2}-9x+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -9 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 3\times 5}}{2\times 3}$$

Square -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\times 3\times 5}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-9\right)±\sqrt{81-12\times 5}}{2\times 3}$$

Multiply -12 times 5.

$$x=\frac{-\left(-9\right)±\sqrt{81-60}}{2\times 3}$$

Add 81 to -60.

$$x=\frac{-\left(-9\right)±\sqrt{21}}{2\times 3}$$

The opposite of -9 is 9.

$$x=\frac{9±\sqrt{21}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{9±\sqrt{21}}{6}$$

Now solve the equation x=\frac{9±\sqrt{21}}{6} when ± is plus. Add 9 to \sqrt{21}.

$$x=\frac{\sqrt{21}+9}{6}$$

Divide 9+\sqrt{21} by 6.

$$x=\frac{\sqrt{21}}{6}+\frac{3}{2}$$

Now solve the equation x=\frac{9±\sqrt{21}}{6} when ± is minus. Subtract \sqrt{21} from 9.

$$x=\frac{9-\sqrt{21}}{6}$$

Divide 9-\sqrt{21} by 6.

$$x=-\frac{\sqrt{21}}{6}+\frac{3}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{21}}{6}+\frac{3}{2} x=-\frac{\sqrt{21}}{6}+\frac{3}{2}$$

Steps for Completing the Square

Subtract 9x from both sides.

$$3x^{2}-9x=-5$$

Divide both sides by 3.

$$\frac{3x^{2}-9x}{3}=-\frac{5}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\left(-\frac{9}{3}\right)x=-\frac{5}{3}$$

Divide -9 by 3.

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

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.

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

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

$$x^{2}-3x+\frac{9}{4}=-\frac{5}{3}+\frac{9}{4}$$

Add -\frac{5}{3} to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-3x+\frac{9}{4}=\frac{7}{12}$$

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}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{7}{12}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{7}{12}}$$

Simplify.

$$x-\frac{3}{2}=\frac{\sqrt{21}}{6} x-\frac{3}{2}=-\frac{\sqrt{21}}{6}$$

Add \frac{3}{2} to both sides of the equation.

$$x=\frac{\sqrt{21}}{6}+\frac{3}{2} x=-\frac{\sqrt{21}}{6}+\frac{3}{2}$$