Solve (x)=3x^2+6x-24

$$( x ) = 3 x ^ { 2 } + 6 x - 24$$

Solve for x

$$x = \frac{\sqrt{313} - 5}{6} \approx 2.115301002$$

$$x=\frac{-\sqrt{313}-5}{6}\approx -3.781967669$$

Steps Using the Quadratic Formula

Subtract 3x^{2} from both sides.

$$x-3x^{2}=6x-24$$

Subtract 6x from both sides.

$$x-3x^{2}-6x=-24$$

Combine x and -6x to get -5x.

$$-5x-3x^{2}=-24$$

Add 24 to both sides.

$$-5x-3x^{2}+24=0$$

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.

$$-3x^{2}-5x+24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, -5 for b, and 24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-3\right)\times 24}}{2\left(-3\right)}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-3\right)\times 24}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-\left(-5\right)±\sqrt{25+12\times 24}}{2\left(-3\right)}$$

Multiply 12 times 24.

$$x=\frac{-\left(-5\right)±\sqrt{25+288}}{2\left(-3\right)}$$

Add 25 to 288.

$$x=\frac{-\left(-5\right)±\sqrt{313}}{2\left(-3\right)}$$

The opposite of -5 is 5.

$$x=\frac{5±\sqrt{313}}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{5±\sqrt{313}}{-6}$$

Now solve the equation x=\frac{5±\sqrt{313}}{-6} when ± is plus. Add 5 to \sqrt{313}.

$$x=\frac{\sqrt{313}+5}{-6}$$

Divide 5+\sqrt{313} by -6.

$$x=\frac{-\sqrt{313}-5}{6}$$

Now solve the equation x=\frac{5±\sqrt{313}}{-6} when ± is minus. Subtract \sqrt{313} from 5.

$$x=\frac{5-\sqrt{313}}{-6}$$

Divide 5-\sqrt{313} by -6.

$$x=\frac{\sqrt{313}-5}{6}$$

The equation is now solved.

$$x=\frac{-\sqrt{313}-5}{6} x=\frac{\sqrt{313}-5}{6}$$

Steps for Completing the Square

Subtract 3x^{2} from both sides.

$$x-3x^{2}=6x-24$$

Subtract 6x from both sides.

$$x-3x^{2}-6x=-24$$

Combine x and -6x to get -5x.

$$-5x-3x^{2}=-24$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-3x^{2}-5x=-24$$

Divide both sides by -3.

$$\frac{-3x^{2}-5x}{-3}=-\frac{24}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\left(-\frac{5}{-3}\right)x=-\frac{24}{-3}$$

Divide -5 by -3.

$$x^{2}+\frac{5}{3}x=-\frac{24}{-3}$$

Divide -24 by -3.

$$x^{2}+\frac{5}{3}x=8$$

Divide \frac{5}{3}, the coefficient of the x term, by 2 to get \frac{5}{6}. Then add the square of \frac{5}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{5}{3}x+\left(\frac{5}{6}\right)^{2}=8+\left(\frac{5}{6}\right)^{2}$$

Square \frac{5}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{3}x+\frac{25}{36}=8+\frac{25}{36}$$

Add 8 to \frac{25}{36}.

$$x^{2}+\frac{5}{3}x+\frac{25}{36}=\frac{313}{36}$$

Factor x^{2}+\frac{5}{3}x+\frac{25}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{6}\right)^{2}=\frac{313}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{6}\right)^{2}}=\sqrt{\frac{313}{36}}$$

Simplify.

$$x+\frac{5}{6}=\frac{\sqrt{313}}{6} x+\frac{5}{6}=-\frac{\sqrt{313}}{6}$$

Subtract \frac{5}{6} from both sides of the equation.

$$x=\frac{\sqrt{313}-5}{6} x=\frac{-\sqrt{313}-5}{6}$$