Solve 15-3m=1+2m+m^2

$$15 - 3 m = 1 + 2 m + m ^ { 2 }$$

Solve for m

$$m=-7$$

$$m=2$$

Steps Using Factoring By Grouping

Subtract 1 from both sides.

$$15-3m-1=2m+m^{2}$$

Subtract 1 from 15 to get 14.

$$14-3m=2m+m^{2}$$

Subtract 2m from both sides.

$$14-3m-2m=m^{2}$$

Combine -3m and -2m to get -5m.

$$14-5m=m^{2}$$

Subtract m^{2} from both sides.

$$14-5m-m^{2}=0$$

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

$$-m^{2}-5m+14=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -m^{2}+am+bm+14. To find a and b, set up a system to be solved.

$$a+b=-5 ab=-14=-14$$

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. List all such integer pairs that give product -14.

$$1,-14 2,-7$$

Calculate the sum for each pair.

$$1-14=-13 2-7=-5$$

The solution is the pair that gives sum -5.

$$a=2 b=-7$$

Rewrite -m^{2}-5m+14 as \left(-m^{2}+2m\right)+\left(-7m+14\right).

$$\left(-m^{2}+2m\right)+\left(-7m+14\right)$$

Factor out m in the first and 7 in the second group.

$$m\left(-m+2\right)+7\left(-m+2\right)$$

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

$$\left(-m+2\right)\left(m+7\right)$$

To find equation solutions, solve -m+2=0 and m+7=0.

$$m=2 m=-7$$

Steps Using the Quadratic Formula

Subtract 1 from both sides.

$$15-3m-1=2m+m^{2}$$

Subtract 1 from 15 to get 14.

$$14-3m=2m+m^{2}$$

Subtract 2m from both sides.

$$14-3m-2m=m^{2}$$

Combine -3m and -2m to get -5m.

$$14-5m=m^{2}$$

Subtract m^{2} from both sides.

$$14-5m-m^{2}=0$$

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.

$$-m^{2}-5m+14=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -5 for b, and 14 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$m=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-1\right)\times 14}}{2\left(-1\right)}$$

Square -5.

$$m=\frac{-\left(-5\right)±\sqrt{25-4\left(-1\right)\times 14}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$m=\frac{-\left(-5\right)±\sqrt{25+4\times 14}}{2\left(-1\right)}$$

Multiply 4 times 14.

$$m=\frac{-\left(-5\right)±\sqrt{25+56}}{2\left(-1\right)}$$

Add 25 to 56.

$$m=\frac{-\left(-5\right)±\sqrt{81}}{2\left(-1\right)}$$

Take the square root of 81.

$$m=\frac{-\left(-5\right)±9}{2\left(-1\right)}$$

The opposite of -5 is 5.

$$m=\frac{5±9}{2\left(-1\right)}$$

Multiply 2 times -1.

$$m=\frac{5±9}{-2}$$

Now solve the equation m=\frac{5±9}{-2} when ± is plus. Add 5 to 9.

$$m=\frac{14}{-2}$$

Divide 14 by -2.

$$m=-7$$

Now solve the equation m=\frac{5±9}{-2} when ± is minus. Subtract 9 from 5.

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

Divide -4 by -2.

$$m=2$$

The equation is now solved.

$$m=-7 m=2$$

Steps for Completing the Square

Subtract 2m from both sides.

$$15-3m-2m=1+m^{2}$$

Combine -3m and -2m to get -5m.

$$15-5m=1+m^{2}$$

Subtract m^{2} from both sides.

$$15-5m-m^{2}=1$$

Subtract 15 from both sides.

$$-5m-m^{2}=1-15$$

Subtract 15 from 1 to get -14.

$$-5m-m^{2}=-14$$

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.

$$-m^{2}-5m=-14$$

Divide both sides by -1.

$$\frac{-m^{2}-5m}{-1}=-\frac{14}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$m^{2}+\left(-\frac{5}{-1}\right)m=-\frac{14}{-1}$$

Divide -5 by -1.

$$m^{2}+5m=-\frac{14}{-1}$$

Divide -14 by -1.

$$m^{2}+5m=14$$

Divide 5, the coefficient of the x term, by 2 to get \frac{5}{2}. Then add the square of \frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$m^{2}+5m+\left(\frac{5}{2}\right)^{2}=14+\left(\frac{5}{2}\right)^{2}$$

Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$m^{2}+5m+\frac{25}{4}=14+\frac{25}{4}$$

Add 14 to \frac{25}{4}.

$$m^{2}+5m+\frac{25}{4}=\frac{81}{4}$$

Factor m^{2}+5m+\frac{25}{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}.

$$\left(m+\frac{5}{2}\right)^{2}=\frac{81}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(m+\frac{5}{2}\right)^{2}}=\sqrt{\frac{81}{4}}$$

Simplify.

$$m+\frac{5}{2}=\frac{9}{2} m+\frac{5}{2}=-\frac{9}{2}$$

Subtract \frac{5}{2} from both sides of the equation.

$$m=2 m=-7$$