Solve 2*(-3/4a^2+3a)=9/2

$$2 \times ( - \frac { 3 } { 4 } a ^ { 2 } + 3 a ) = \frac { 9 } { 2 }$$

Solve for a

$$a=1$$

$$a=3$$

Steps Using the Quadratic Formula

Divide both sides by 2.

$$-\frac{3}{4}a^{2}+3a=\frac{\frac{9}{2}}{2}$$

Express \frac{\frac{9}{2}}{2} as a single fraction.

$$-\frac{3}{4}a^{2}+3a=\frac{9}{2\times 2}$$

Multiply 2 and 2 to get 4.

$$-\frac{3}{4}a^{2}+3a=\frac{9}{4}$$

Subtract \frac{9}{4} from both sides.

$$-\frac{3}{4}a^{2}+3a-\frac{9}{4}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{3}{4} for a, 3 for b, and -\frac{9}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-3±\sqrt{3^{2}-4\left(-\frac{3}{4}\right)\left(-\frac{9}{4}\right)}}{2\left(-\frac{3}{4}\right)}$$

Square 3.

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

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

$$a=\frac{-3±\sqrt{9+3\left(-\frac{9}{4}\right)}}{2\left(-\frac{3}{4}\right)}$$

Multiply 3 times -\frac{9}{4}.

$$a=\frac{-3±\sqrt{9-\frac{27}{4}}}{2\left(-\frac{3}{4}\right)}$$

Add 9 to -\frac{27}{4}.

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

Take the square root of \frac{9}{4}.

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

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

$$a=\frac{-3±\frac{3}{2}}{-\frac{3}{2}}$$

Now solve the equation a=\frac{-3±\frac{3}{2}}{-\frac{3}{2}} when ± is plus. Add -3 to \frac{3}{2}.

$$a=-\frac{\frac{3}{2}}{-\frac{3}{2}}$$

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

$$a=1$$

Now solve the equation a=\frac{-3±\frac{3}{2}}{-\frac{3}{2}} when ± is minus. Subtract \frac{3}{2} from -3.

$$a=-\frac{\frac{9}{2}}{-\frac{3}{2}}$$

Divide -\frac{9}{2} by -\frac{3}{2} by multiplying -\frac{9}{2} by the reciprocal of -\frac{3}{2}.

$$a=3$$

The equation is now solved.

$$a=1 a=3$$

Steps for Completing the Square

Divide both sides by 2.

$$-\frac{3}{4}a^{2}+3a=\frac{\frac{9}{2}}{2}$$

Express \frac{\frac{9}{2}}{2} as a single fraction.

$$-\frac{3}{4}a^{2}+3a=\frac{9}{2\times 2}$$

Multiply 2 and 2 to get 4.

$$-\frac{3}{4}a^{2}+3a=\frac{9}{4}$$

Divide both sides of the equation by -\frac{3}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-\frac{3}{4}a^{2}+3a}{-\frac{3}{4}}=\frac{\frac{9}{4}}{-\frac{3}{4}}$$

Dividing by -\frac{3}{4} undoes the multiplication by -\frac{3}{4}.

$$a^{2}+\frac{3}{-\frac{3}{4}}a=\frac{\frac{9}{4}}{-\frac{3}{4}}$$

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

$$a^{2}-4a=\frac{\frac{9}{4}}{-\frac{3}{4}}$$

Divide \frac{9}{4} by -\frac{3}{4} by multiplying \frac{9}{4} by the reciprocal of -\frac{3}{4}.

$$a^{2}-4a=-3$$

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

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

Square -2.

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

Add -3 to 4.

$$a^{2}-4a+4=1$$

Factor a^{2}-4a+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}.

$$\left(a-2\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-2\right)^{2}}=\sqrt{1}$$

Simplify.

$$a-2=1 a-2=-1$$

Add 2 to both sides of the equation.

$$a=3 a=1$$