Solve 2*(-3/4a^2+3a)=6

$$2 \times ( - \frac { 3 } { 4 } a ^ { 2 } + 3 a ) = 6$$

Solve for a

$$a=2$$

Steps Using the Quadratic Formula

Divide both sides by 2.

$$-\frac{3}{4}a^{2}+3a=\frac{6}{2}$$

Divide 6 by 2 to get 3.

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

Subtract 3 from both sides.

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

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

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

Square 3.

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

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

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

Multiply 3 times -3.

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

Add 9 to -9.

$$a=\frac{-3±\sqrt{0}}{2\left(-\frac{3}{4}\right)}$$

Take the square root of 0.

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

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

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

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

$$a=2$$

Steps for Completing the Square

Divide both sides by 2.

$$-\frac{3}{4}a^{2}+3a=\frac{6}{2}$$

Divide 6 by 2 to get 3.

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

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

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

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

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

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

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

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

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

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.

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

Square -2.

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

Add -4 to 4.

$$a^{2}-4a+4=0$$

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}.

$$\left(a-2\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-2\right)^{2}}=\sqrt{0}$$

Simplify.

$$a-2=0 a-2=0$$

Add 2 to both sides of the equation.

$$a=2 a=2$$

The equation is now solved. Solutions are the same.

$$a=2$$