Solve 3x(x+1)=215

$$3x(x+1)=215$$

Solve for x

$$x=\frac{\sqrt{2589}}{6}-\frac{1}{2}\approx 7.980369489$$

$$x=-\frac{\sqrt{2589}}{6}-\frac{1}{2}\approx -8.980369489$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3x by x+1.

$$3x^{2}+3x=215$$

Subtract 215 from both sides.

$$3x^{2}+3x-215=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 3 for b, and -215 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\times 3\left(-215\right)}}{2\times 3}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\times 3\left(-215\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-3±\sqrt{9-12\left(-215\right)}}{2\times 3}$$

Multiply -12 times -215.

$$x=\frac{-3±\sqrt{9+2580}}{2\times 3}$$

Add 9 to 2580.

$$x=\frac{-3±\sqrt{2589}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-3±\sqrt{2589}}{6}$$

Now solve the equation x=\frac{-3±\sqrt{2589}}{6} when ± is plus. Add -3 to \sqrt{2589}.

$$x=\frac{\sqrt{2589}-3}{6}$$

Divide -3+\sqrt{2589} by 6.

$$x=\frac{\sqrt{2589}}{6}-\frac{1}{2}$$

Now solve the equation x=\frac{-3±\sqrt{2589}}{6} when ± is minus. Subtract \sqrt{2589} from -3.

$$x=\frac{-\sqrt{2589}-3}{6}$$

Divide -3-\sqrt{2589} by 6.

$$x=-\frac{\sqrt{2589}}{6}-\frac{1}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{2589}}{6}-\frac{1}{2} x=-\frac{\sqrt{2589}}{6}-\frac{1}{2}$$

Steps for Completing the Square

Use the distributive property to multiply 3x by x+1.

$$3x^{2}+3x=215$$

Divide both sides by 3.

$$\frac{3x^{2}+3x}{3}=\frac{215}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{3}{3}x=\frac{215}{3}$$

Divide 3 by 3.

$$x^{2}+x=\frac{215}{3}$$

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

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=\frac{215}{3}+\left(\frac{1}{2}\right)^{2}$$

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+x+\frac{1}{4}=\frac{215}{3}+\frac{1}{4}$$

Add \frac{215}{3} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+x+\frac{1}{4}=\frac{863}{12}$$

Factor x^{2}+x+\frac{1}{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}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{863}{12}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{863}{12}}$$

Simplify.

$$x+\frac{1}{2}=\frac{\sqrt{2589}}{6} x+\frac{1}{2}=-\frac{\sqrt{2589}}{6}$$

Subtract \frac{1}{2} from both sides of the equation.

$$x=\frac{\sqrt{2589}}{6}-\frac{1}{2} x=-\frac{\sqrt{2589}}{6}-\frac{1}{2}$$