Solve m^2=28+3m

$$m ^ { 2 } = 28 + 3 m$$

Solve for m

$$m=-4$$

$$m=7$$

Steps Using Factoring

Subtract 28 from both sides.

$$m^{2}-28=3m$$

Subtract 3m from both sides.

$$m^{2}-28-3m=0$$

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

$$m^{2}-3m-28=0$$

To solve the equation, factor m^{2}-3m-28 using formula m^{2}+\left(a+b\right)m+ab=\left(m+a\right)\left(m+b\right). To find a and b, set up a system to be solved.

$$a+b=-3 ab=-28$$

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 -28.

$$1,-28 2,-14 4,-7$$

Calculate the sum for each pair.

$$1-28=-27 2-14=-12 4-7=-3$$

The solution is the pair that gives sum -3.

$$a=-7 b=4$$

Rewrite factored expression \left(m+a\right)\left(m+b\right) using the obtained values.

$$\left(m-7\right)\left(m+4\right)$$

To find equation solutions, solve m-7=0 and m+4=0.

$$m=7 m=-4$$

Steps Using Factoring By Grouping

Subtract 28 from both sides.

$$m^{2}-28=3m$$

Subtract 3m from both sides.

$$m^{2}-28-3m=0$$

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

$$m^{2}-3m-28=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as m^{2}+am+bm-28. To find a and b, set up a system to be solved.

$$a+b=-3 ab=1\left(-28\right)=-28$$

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 -28.

$$1,-28 2,-14 4,-7$$

Calculate the sum for each pair.

$$1-28=-27 2-14=-12 4-7=-3$$

The solution is the pair that gives sum -3.

$$a=-7 b=4$$

Rewrite m^{2}-3m-28 as \left(m^{2}-7m\right)+\left(4m-28\right).

$$\left(m^{2}-7m\right)+\left(4m-28\right)$$

Factor out m in the first and 4 in the second group.

$$m\left(m-7\right)+4\left(m-7\right)$$

Factor out common term m-7 by using distributive property.

$$\left(m-7\right)\left(m+4\right)$$

To find equation solutions, solve m-7=0 and m+4=0.

$$m=7 m=-4$$

Steps Using the Quadratic Formula

Subtract 28 from both sides.

$$m^{2}-28=3m$$

Subtract 3m from both sides.

$$m^{2}-28-3m=0$$

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.

$$m^{2}-3m-28=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -3 for b, and -28 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$m=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-28\right)}}{2}$$

Square -3.

$$m=\frac{-\left(-3\right)±\sqrt{9-4\left(-28\right)}}{2}$$

Multiply -4 times -28.

$$m=\frac{-\left(-3\right)±\sqrt{9+112}}{2}$$

Add 9 to 112.

$$m=\frac{-\left(-3\right)±\sqrt{121}}{2}$$

Take the square root of 121.

$$m=\frac{-\left(-3\right)±11}{2}$$

The opposite of -3 is 3.

$$m=\frac{3±11}{2}$$

Now solve the equation m=\frac{3±11}{2} when ± is plus. Add 3 to 11.

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

Divide 14 by 2.

$$m=7$$

Now solve the equation m=\frac{3±11}{2} when ± is minus. Subtract 11 from 3.

$$m=-\frac{8}{2}$$

Divide -8 by 2.

$$m=-4$$

The equation is now solved.

$$m=7 m=-4$$

Steps for Completing the Square

Subtract 3m from both sides.

$$m^{2}-3m=28$$

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.

$$m^{2}-3m+\left(-\frac{3}{2}\right)^{2}=28+\left(-\frac{3}{2}\right)^{2}$$

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

$$m^{2}-3m+\frac{9}{4}=28+\frac{9}{4}$$

Add 28 to \frac{9}{4}.

$$m^{2}-3m+\frac{9}{4}=\frac{121}{4}$$

Factor m^{2}-3m+\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}.

$$\left(m-\frac{3}{2}\right)^{2}=\frac{121}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(m-\frac{3}{2}\right)^{2}}=\sqrt{\frac{121}{4}}$$

Simplify.

$$m-\frac{3}{2}=\frac{11}{2} m-\frac{3}{2}=-\frac{11}{2}$$

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

$$m=7 m=-4$$