Solve 3-h=h^2-2/h

$$3 - h = \frac { h ^ { 2 } - 2 } { h }$$

Solve for h

$$h=-\frac{1}{2}=-0.5$$

$$h=2$$

Steps Using Factoring By Grouping

Variable h cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by h.

$$h\times 3-hh=h^{2}-2$$

Multiply h and h to get h^{2}.

$$h\times 3-h^{2}=h^{2}-2$$

Subtract h^{2} from both sides.

$$h\times 3-h^{2}-h^{2}=-2$$

Combine -h^{2} and -h^{2} to get -2h^{2}.

$$h\times 3-2h^{2}=-2$$

Add 2 to both sides.

$$h\times 3-2h^{2}+2=0$$

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

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

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

$$a+b=3 ab=-2\times 2=-4$$

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. List all such integer pairs that give product -4.

$$-1,4 -2,2$$

Calculate the sum for each pair.

$$-1+4=3 -2+2=0$$

The solution is the pair that gives sum 3.

$$a=4 b=-1$$

Rewrite -2h^{2}+3h+2 as \left(-2h^{2}+4h\right)+\left(-h+2\right).

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

Factor out 2h in -2h^{2}+4h.

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

Factor out common term -h+2 by using distributive property.

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

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

$$h=2 h=-\frac{1}{2}$$

Steps Using the Quadratic Formula

Variable h cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by h.

$$h\times 3-hh=h^{2}-2$$

Multiply h and h to get h^{2}.

$$h\times 3-h^{2}=h^{2}-2$$

Subtract h^{2} from both sides.

$$h\times 3-h^{2}-h^{2}=-2$$

Combine -h^{2} and -h^{2} to get -2h^{2}.

$$h\times 3-2h^{2}=-2$$

Add 2 to both sides.

$$h\times 3-2h^{2}+2=0$$

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.

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

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

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

Square 3.

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

Multiply -4 times -2.

$$h=\frac{-3±\sqrt{9+8\times 2}}{2\left(-2\right)}$$

Multiply 8 times 2.

$$h=\frac{-3±\sqrt{9+16}}{2\left(-2\right)}$$

Add 9 to 16.

$$h=\frac{-3±\sqrt{25}}{2\left(-2\right)}$$

Take the square root of 25.

$$h=\frac{-3±5}{2\left(-2\right)}$$

Multiply 2 times -2.

$$h=\frac{-3±5}{-4}$$

Now solve the equation h=\frac{-3±5}{-4} when ± is plus. Add -3 to 5.

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

Reduce the fraction \frac{2}{-4} to lowest terms by extracting and canceling out 2.

$$h=-\frac{1}{2}$$

Now solve the equation h=\frac{-3±5}{-4} when ± is minus. Subtract 5 from -3.

$$h=-\frac{8}{-4}$$

Divide -8 by -4.

$$h=2$$

The equation is now solved.

$$h=-\frac{1}{2} h=2$$

Steps for Completing the Square

Variable h cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by h.

$$h\times 3-hh=h^{2}-2$$

Multiply h and h to get h^{2}.

$$h\times 3-h^{2}=h^{2}-2$$

Subtract h^{2} from both sides.

$$h\times 3-h^{2}-h^{2}=-2$$

Combine -h^{2} and -h^{2} to get -2h^{2}.

$$h\times 3-2h^{2}=-2$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-2h^{2}+3h=-2$$

Divide both sides by -2.

$$\frac{-2h^{2}+3h}{-2}=-\frac{2}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$h^{2}+\frac{3}{-2}h=-\frac{2}{-2}$$

Divide 3 by -2.

$$h^{2}-\frac{3}{2}h=-\frac{2}{-2}$$

Divide -2 by -2.

$$h^{2}-\frac{3}{2}h=1$$

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.

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

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

$$h^{2}-\frac{3}{2}h+\frac{9}{16}=1+\frac{9}{16}$$

Add 1 to \frac{9}{16}.

$$h^{2}-\frac{3}{2}h+\frac{9}{16}=\frac{25}{16}$$

Factor h^{2}-\frac{3}{2}h+\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}.

$$\left(h-\frac{3}{4}\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(h-\frac{3}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

$$h-\frac{3}{4}=\frac{5}{4} h-\frac{3}{4}=-\frac{5}{4}$$

Add \frac{3}{4} to both sides of the equation.

$$h=2 h=-\frac{1}{2}$$