Solve -1+a^2-3/2a=-1/2

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

Solve for a

$$a=-1$$

$$a=2$$

Steps Using Factoring By Grouping

Variable a cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 2a, the least common multiple of 2a,2.

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

Subtract 3 from 1 to get -2.

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

To find the opposite of -2+a^{2}, find the opposite of each term.

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

Add a to both sides.

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

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

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.

$$a+b=1 ab=-2=-2$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=2 b=-1$$

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

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

Factor out -a in the first and -1 in the second group.

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

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

Variable a cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 2a, the least common multiple of 2a,2.

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

Subtract 3 from 1 to get -2.

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

To find the opposite of -2+a^{2}, find the opposite of each term.

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

Add a to both sides.

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

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.

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

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

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

Square 1.

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

Multiply -4 times -1.

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

Multiply 4 times 2.

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

Add 1 to 8.

$$a=\frac{-1±\sqrt{9}}{2\left(-1\right)}$$

Take the square root of 9.

$$a=\frac{-1±3}{2\left(-1\right)}$$

Multiply 2 times -1.

$$a=\frac{-1±3}{-2}$$

Now solve the equation a=\frac{-1±3}{-2} when ± is plus. Add -1 to 3.

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

Divide 2 by -2.

$$a=-1$$

Now solve the equation a=\frac{-1±3}{-2} when ± is minus. Subtract 3 from -1.

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

Divide -4 by -2.

$$a=2$$

The equation is now solved.

$$a=-1 a=2$$

Steps for Completing the Square

Variable a cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 2a, the least common multiple of 2a,2.

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

Subtract 3 from 1 to get -2.

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

To find the opposite of -2+a^{2}, find the opposite of each term.

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

Add a to both sides.

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

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide 1 by -1.

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

Divide -2 by -1.

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

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

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

Add 2 to \frac{1}{4}.

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

Factor a^{2}-a+\frac{1}{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}.

$$\left(a-\frac{1}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-\frac{1}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

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

Add \frac{1}{2} to both sides of the equation.

$$a=2 a=-1$$