Solve -3/2a^2+9a/2=3

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

Solve for a

$$a=1$$

$$a=2$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 2.

$$-3a^{2}+9a=6$$

Subtract 6 from both sides.

$$-3a^{2}+9a-6=0$$

Divide both sides by 3.

$$-a^{2}+3a-2=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -a^{2}+aa+ba-2. To find a and b, set up a system to be solved.

$$a+b=3 ab=-\left(-2\right)=2$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=2 b=1$$

Rewrite -a^{2}+3a-2 as \left(-a^{2}+2a\right)+\left(a-2\right).

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

Factor out -a in -a^{2}+2a.

$$-a\left(a-2\right)+a-2$$

Factor out common term a-2 by using distributive property.

$$\left(a-2\right)\left(-a+1\right)$$

To find equation solutions, solve a-2=0 and -a+1=0.

$$a=2 a=1$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

$$-3a^{2}+9a=6$$

Subtract 6 from both sides.

$$-3a^{2}+9a-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 9 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-9±\sqrt{9^{2}-4\left(-3\right)\left(-6\right)}}{2\left(-3\right)}$$

Square 9.

$$a=\frac{-9±\sqrt{81-4\left(-3\right)\left(-6\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$a=\frac{-9±\sqrt{81+12\left(-6\right)}}{2\left(-3\right)}$$

Multiply 12 times -6.

$$a=\frac{-9±\sqrt{81-72}}{2\left(-3\right)}$$

Add 81 to -72.

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

Take the square root of 9.

$$a=\frac{-9±3}{2\left(-3\right)}$$

Multiply 2 times -3.

$$a=\frac{-9±3}{-6}$$

Now solve the equation a=\frac{-9±3}{-6} when ± is plus. Add -9 to 3.

$$a=-\frac{6}{-6}$$

Divide -6 by -6.

$$a=1$$

Now solve the equation a=\frac{-9±3}{-6} when ± is minus. Subtract 3 from -9.

$$a=-\frac{12}{-6}$$

Divide -12 by -6.

$$a=2$$

The equation is now solved.

$$a=1 a=2$$

Steps for Completing the Square

Multiply both sides of the equation by 2.

$$-3a^{2}+9a=6$$

Divide both sides by -3.

$$\frac{-3a^{2}+9a}{-3}=\frac{6}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$a^{2}+\frac{9}{-3}a=\frac{6}{-3}$$

Divide 9 by -3.

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

Divide 6 by -3.

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

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

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

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

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

Add -2 to \frac{9}{4}.

$$a^{2}-3a+\frac{9}{4}=\frac{1}{4}$$

Factor a^{2}-3a+\frac{9}{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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$a-\frac{3}{2}=\frac{1}{2} a-\frac{3}{2}=-\frac{1}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$a=2 a=1$$