Solve 3x^2-9x=x-5

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

Solve for x

$$x = \frac{\sqrt{10} + 5}{3} \approx 2.72075922$$

$$x=\frac{5-\sqrt{10}}{3}\approx 0.612574113$$

Steps Using the Quadratic Formula

Subtract x from both sides.

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

Combine -9x and -x to get -10x.

$$3x^{2}-10x=-5$$

Add 5 to both sides.

$$3x^{2}-10x+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -10 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 3\times 5}}{2\times 3}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 3\times 5}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-10\right)±\sqrt{100-12\times 5}}{2\times 3}$$

Multiply -12 times 5.

$$x=\frac{-\left(-10\right)±\sqrt{100-60}}{2\times 3}$$

Add 100 to -60.

$$x=\frac{-\left(-10\right)±\sqrt{40}}{2\times 3}$$

Take the square root of 40.

$$x=\frac{-\left(-10\right)±2\sqrt{10}}{2\times 3}$$

The opposite of -10 is 10.

$$x=\frac{10±2\sqrt{10}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{10±2\sqrt{10}}{6}$$

Now solve the equation x=\frac{10±2\sqrt{10}}{6} when ± is plus. Add 10 to 2\sqrt{10}.

$$x=\frac{2\sqrt{10}+10}{6}$$

Divide 10+2\sqrt{10} by 6.

$$x=\frac{\sqrt{10}+5}{3}$$

Now solve the equation x=\frac{10±2\sqrt{10}}{6} when ± is minus. Subtract 2\sqrt{10} from 10.

$$x=\frac{10-2\sqrt{10}}{6}$$

Divide 10-2\sqrt{10} by 6.

$$x=\frac{5-\sqrt{10}}{3}$$

The equation is now solved.

$$x=\frac{\sqrt{10}+5}{3} x=\frac{5-\sqrt{10}}{3}$$

Steps for Completing the Square

Subtract x from both sides.

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

Combine -9x and -x to get -10x.

$$3x^{2}-10x=-5$$

Divide both sides by 3.

$$\frac{3x^{2}-10x}{3}=-\frac{5}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{10}{3}x=-\frac{5}{3}$$

Divide -\frac{10}{3}, the coefficient of the x term, by 2 to get -\frac{5}{3}. Then add the square of -\frac{5}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{10}{3}x+\left(-\frac{5}{3}\right)^{2}=-\frac{5}{3}+\left(-\frac{5}{3}\right)^{2}$$

Square -\frac{5}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{10}{3}x+\frac{25}{9}=-\frac{5}{3}+\frac{25}{9}$$

Add -\frac{5}{3} to \frac{25}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{10}{3}x+\frac{25}{9}=\frac{10}{9}$$

Factor x^{2}-\frac{10}{3}x+\frac{25}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{3}\right)^{2}=\frac{10}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{3}\right)^{2}}=\sqrt{\frac{10}{9}}$$

Simplify.

$$x-\frac{5}{3}=\frac{\sqrt{10}}{3} x-\frac{5}{3}=-\frac{\sqrt{10}}{3}$$

Add \frac{5}{3} to both sides of the equation.

$$x=\frac{\sqrt{10}+5}{3} x=\frac{5-\sqrt{10}}{3}$$