Solve 2/3(a+2)^2-5/3(a+2)-4=2

$$\frac { 2 } { 3 } ( a + 2 ) ^ { 2 } - \frac { 5 } { 3 } ( a + 2 ) - 4 = 2$$

Solve for a

$$a=-4$$

$$a = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(a+2\right)^{2}.

$$\frac{2}{3}\left(a^{2}+4a+4\right)-\frac{5}{3}\left(a+2\right)-4=2$$

Use the distributive property to multiply \frac{2}{3} by a^{2}+4a+4.

$$\frac{2}{3}a^{2}+\frac{8}{3}a+\frac{8}{3}-\frac{5}{3}\left(a+2\right)-4=2$$

Use the distributive property to multiply -\frac{5}{3} by a+2.

$$\frac{2}{3}a^{2}+\frac{8}{3}a+\frac{8}{3}-\frac{5}{3}a-\frac{10}{3}-4=2$$

Combine \frac{8}{3}a and -\frac{5}{3}a to get a.

$$\frac{2}{3}a^{2}+a+\frac{8}{3}-\frac{10}{3}-4=2$$

Subtract \frac{10}{3} from \frac{8}{3} to get -\frac{2}{3}.

$$\frac{2}{3}a^{2}+a-\frac{2}{3}-4=2$$

Subtract 4 from -\frac{2}{3} to get -\frac{14}{3}.

$$\frac{2}{3}a^{2}+a-\frac{14}{3}=2$$

Subtract 2 from both sides.

$$\frac{2}{3}a^{2}+a-\frac{14}{3}-2=0$$

Subtract 2 from -\frac{14}{3} to get -\frac{20}{3}.

$$\frac{2}{3}a^{2}+a-\frac{20}{3}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{2}{3} for a, 1 for b, and -\frac{20}{3} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-1±\sqrt{1^{2}-4\times \frac{2}{3}\left(-\frac{20}{3}\right)}}{2\times \frac{2}{3}}$$

Square 1.

$$a=\frac{-1±\sqrt{1-4\times \frac{2}{3}\left(-\frac{20}{3}\right)}}{2\times \frac{2}{3}}$$

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

$$a=\frac{-1±\sqrt{1-\frac{8}{3}\left(-\frac{20}{3}\right)}}{2\times \frac{2}{3}}$$

Multiply -\frac{8}{3} times -\frac{20}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$a=\frac{-1±\sqrt{1+\frac{160}{9}}}{2\times \frac{2}{3}}$$

Add 1 to \frac{160}{9}.

$$a=\frac{-1±\sqrt{\frac{169}{9}}}{2\times \frac{2}{3}}$$

Take the square root of \frac{169}{9}.

$$a=\frac{-1±\frac{13}{3}}{2\times \frac{2}{3}}$$

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

$$a=\frac{-1±\frac{13}{3}}{\frac{4}{3}}$$

Now solve the equation a=\frac{-1±\frac{13}{3}}{\frac{4}{3}} when ± is plus. Add -1 to \frac{13}{3}.

$$a=\frac{\frac{10}{3}}{\frac{4}{3}}$$

Divide \frac{10}{3} by \frac{4}{3} by multiplying \frac{10}{3} by the reciprocal of \frac{4}{3}.

$$a=\frac{5}{2}$$

Now solve the equation a=\frac{-1±\frac{13}{3}}{\frac{4}{3}} when ± is minus. Subtract \frac{13}{3} from -1.

$$a=-\frac{\frac{16}{3}}{\frac{4}{3}}$$

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

$$a=-4$$

The equation is now solved.

$$a=\frac{5}{2} a=-4$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(a+2\right)^{2}.

$$\frac{2}{3}\left(a^{2}+4a+4\right)-\frac{5}{3}\left(a+2\right)-4=2$$

Use the distributive property to multiply \frac{2}{3} by a^{2}+4a+4.

$$\frac{2}{3}a^{2}+\frac{8}{3}a+\frac{8}{3}-\frac{5}{3}\left(a+2\right)-4=2$$

Use the distributive property to multiply -\frac{5}{3} by a+2.

$$\frac{2}{3}a^{2}+\frac{8}{3}a+\frac{8}{3}-\frac{5}{3}a-\frac{10}{3}-4=2$$

Combine \frac{8}{3}a and -\frac{5}{3}a to get a.

$$\frac{2}{3}a^{2}+a+\frac{8}{3}-\frac{10}{3}-4=2$$

Subtract \frac{10}{3} from \frac{8}{3} to get -\frac{2}{3}.

$$\frac{2}{3}a^{2}+a-\frac{2}{3}-4=2$$

Subtract 4 from -\frac{2}{3} to get -\frac{14}{3}.

$$\frac{2}{3}a^{2}+a-\frac{14}{3}=2$$

Add \frac{14}{3} to both sides.

$$\frac{2}{3}a^{2}+a=2+\frac{14}{3}$$

Add 2 and \frac{14}{3} to get \frac{20}{3}.

$$\frac{2}{3}a^{2}+a=\frac{20}{3}$$

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

$$\frac{\frac{2}{3}a^{2}+a}{\frac{2}{3}}=\frac{\frac{20}{3}}{\frac{2}{3}}$$

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

$$a^{2}+\frac{1}{\frac{2}{3}}a=\frac{\frac{20}{3}}{\frac{2}{3}}$$

Divide 1 by \frac{2}{3} by multiplying 1 by the reciprocal of \frac{2}{3}.

$$a^{2}+\frac{3}{2}a=\frac{\frac{20}{3}}{\frac{2}{3}}$$

Divide \frac{20}{3} by \frac{2}{3} by multiplying \frac{20}{3} by the reciprocal of \frac{2}{3}.

$$a^{2}+\frac{3}{2}a=10$$

Divide \frac{3}{2}, the coefficient of the x term, by 2 to get \frac{3}{4}. Then add the square of \frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}+\frac{3}{2}a+\left(\frac{3}{4}\right)^{2}=10+\left(\frac{3}{4}\right)^{2}$$

Square \frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$a^{2}+\frac{3}{2}a+\frac{9}{16}=10+\frac{9}{16}$$

Add 10 to \frac{9}{16}.

$$a^{2}+\frac{3}{2}a+\frac{9}{16}=\frac{169}{16}$$

Factor a^{2}+\frac{3}{2}a+\frac{9}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(a+\frac{3}{4}\right)^{2}=\frac{169}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a+\frac{3}{4}\right)^{2}}=\sqrt{\frac{169}{16}}$$

Simplify.

$$a+\frac{3}{4}=\frac{13}{4} a+\frac{3}{4}=-\frac{13}{4}$$

Subtract \frac{3}{4} from both sides of the equation.

$$a=\frac{5}{2} a=-4$$