Solve -h^2+3h+1=4h-1

$$- h ^ { 2 } + 3 h + 1 = 4 h - 1$$

Solve for h

$$h=-2$$

$$h=1$$

Steps Using Factoring By Grouping

Subtract 4h from both sides.

$$-h^{2}+3h+1-4h=-1$$

Combine 3h and -4h to get -h.

$$-h^{2}-h+1=-1$$

Add 1 to both sides.

$$-h^{2}-h+1+1=0$$

Add 1 and 1 to get 2.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -h^{2}+ah+bh+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 negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=1 b=-2$$

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

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

Factor out h in the first and 2 in the second group.

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

Factor out common term -h+1 by using distributive property.

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

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

$$h=1 h=-2$$

Steps Using the Quadratic Formula

Subtract 4h from both sides.

$$-h^{2}+3h+1-4h=-1$$

Combine 3h and -4h to get -h.

$$-h^{2}-h+1=-1$$

Add 1 to both sides.

$$-h^{2}-h+1+1=0$$

Add 1 and 1 to get 2.

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

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}.

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

Multiply -4 times -1.

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

Multiply 4 times 2.

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

Add 1 to 8.

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

Take the square root of 9.

$$h=\frac{-\left(-1\right)±3}{2\left(-1\right)}$$

The opposite of -1 is 1.

$$h=\frac{1±3}{2\left(-1\right)}$$

Multiply 2 times -1.

$$h=\frac{1±3}{-2}$$

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

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

Divide 4 by -2.

$$h=-2$$

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

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

Divide -2 by -2.

$$h=1$$

The equation is now solved.

$$h=-2 h=1$$

Steps for Completing the Square

Subtract 4h from both sides.

$$-h^{2}+3h+1-4h=-1$$

Combine 3h and -4h to get -h.

$$-h^{2}-h+1=-1$$

Subtract 1 from both sides.

$$-h^{2}-h=-1-1$$

Subtract 1 from -1 to get -2.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide -1 by -1.

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

Divide -2 by -1.

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

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.

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

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

$$h^{2}+h+\frac{1}{4}=2+\frac{1}{4}$$

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

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

Factor h^{2}+h+\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}.

$$\left(h+\frac{1}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract \frac{1}{2} from both sides of the equation.

$$h=1 h=-2$$