Solve 2/3x^2-6x-24=0

$$\frac { 2 } { 3 } x ^ { 2 } - 6 x - 24 = 0$$

Solve for x

$$x=-3$$

$$x=12$$

Steps Using the Quadratic Formula

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.

$$\frac{2}{3}x^{2}-6x-24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{2}{3} for a, -6 for b, and -24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times \frac{2}{3}\left(-24\right)}}{2\times \frac{2}{3}}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times \frac{2}{3}\left(-24\right)}}{2\times \frac{2}{3}}$$

Multiply -4 times \frac{2}{3}.

$$x=\frac{-\left(-6\right)±\sqrt{36-\frac{8}{3}\left(-24\right)}}{2\times \frac{2}{3}}$$

Multiply -\frac{8}{3} times -24.

$$x=\frac{-\left(-6\right)±\sqrt{36+64}}{2\times \frac{2}{3}}$$

Add 36 to 64.

$$x=\frac{-\left(-6\right)±\sqrt{100}}{2\times \frac{2}{3}}$$

Take the square root of 100.

$$x=\frac{-\left(-6\right)±10}{2\times \frac{2}{3}}$$

The opposite of -6 is 6.

$$x=\frac{6±10}{2\times \frac{2}{3}}$$

Multiply 2 times \frac{2}{3}.

$$x=\frac{6±10}{\frac{4}{3}}$$

Now solve the equation x=\frac{6±10}{\frac{4}{3}} when ± is plus. Add 6 to 10.

$$x=\frac{16}{\frac{4}{3}}$$

Divide 16 by \frac{4}{3} by multiplying 16 by the reciprocal of \frac{4}{3}.

$$x=12$$

Now solve the equation x=\frac{6±10}{\frac{4}{3}} when ± is minus. Subtract 10 from 6.

$$x=-\frac{4}{\frac{4}{3}}$$

Divide -4 by \frac{4}{3} by multiplying -4 by the reciprocal of \frac{4}{3}.

$$x=-3$$

The equation is now solved.

$$x=12 x=-3$$

Steps for Completing the Square

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.

$$\frac{2}{3}x^{2}-6x-24=0$$

Add 24 to both sides of the equation.

$$\frac{2}{3}x^{2}-6x-24-\left(-24\right)=-\left(-24\right)$$

Subtracting -24 from itself leaves 0.

$$\frac{2}{3}x^{2}-6x=-\left(-24\right)$$

Subtract -24 from 0.

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

Divide both sides of the equation by \frac{2}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

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

Dividing by \frac{2}{3} undoes the multiplication by \frac{2}{3}.

$$x^{2}+\left(-\frac{6}{\frac{2}{3}}\right)x=\frac{24}{\frac{2}{3}}$$

Divide -6 by \frac{2}{3} by multiplying -6 by the reciprocal of \frac{2}{3}.

$$x^{2}-9x=\frac{24}{\frac{2}{3}}$$

Divide 24 by \frac{2}{3} by multiplying 24 by the reciprocal of \frac{2}{3}.

$$x^{2}-9x=36$$

Divide -9, the coefficient of the x term, by 2 to get -\frac{9}{2}. Then add the square of -\frac{9}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-9x+\left(-\frac{9}{2}\right)^{2}=36+\left(-\frac{9}{2}\right)^{2}$$

Square -\frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-9x+\frac{81}{4}=36+\frac{81}{4}$$

Add 36 to \frac{81}{4}.

$$x^{2}-9x+\frac{81}{4}=\frac{225}{4}$$

Factor x^{2}-9x+\frac{81}{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}.

$$\left(x-\frac{9}{2}\right)^{2}=\frac{225}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{2}\right)^{2}}=\sqrt{\frac{225}{4}}$$

Simplify.

$$x-\frac{9}{2}=\frac{15}{2} x-\frac{9}{2}=-\frac{15}{2}$$

Add \frac{9}{2} to both sides of the equation.

$$x=12 x=-3$$