Solve 5x+72=x^2+6x

$$5 x + 72 = x ^ { 2 } + 6 x$$

Solve for x

$$x=-9$$

$$x=8$$

Steps Using Factoring By Grouping

Subtract x^{2} from both sides.

$$5x+72-x^{2}=6x$$

Subtract 6x from both sides.

$$5x+72-x^{2}-6x=0$$

Combine 5x and -6x to get -x.

$$-x+72-x^{2}=0$$

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

$$-x^{2}-x+72=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+72. To find a and b, set up a system to be solved.

$$a+b=-1 ab=-72=-72$$

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 -72.

$$1,-72 2,-36 3,-24 4,-18 6,-12 8,-9$$

Calculate the sum for each pair.

$$1-72=-71 2-36=-34 3-24=-21 4-18=-14 6-12=-6 8-9=-1$$

The solution is the pair that gives sum -1.

$$a=8 b=-9$$

Rewrite -x^{2}-x+72 as \left(-x^{2}+8x\right)+\left(-9x+72\right).

$$\left(-x^{2}+8x\right)+\left(-9x+72\right)$$

Factor out x in the first and 9 in the second group.

$$x\left(-x+8\right)+9\left(-x+8\right)$$

Factor out common term -x+8 by using distributive property.

$$\left(-x+8\right)\left(x+9\right)$$

To find equation solutions, solve -x+8=0 and x+9=0.

$$x=8 x=-9$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

$$5x+72-x^{2}=6x$$

Subtract 6x from both sides.

$$5x+72-x^{2}-6x=0$$

Combine 5x and -6x to get -x.

$$-x+72-x^{2}=0$$

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.

$$-x^{2}-x+72=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -1 for b, and 72 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-1\right)\times 72}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-1\right)±\sqrt{1+4\times 72}}{2\left(-1\right)}$$

Multiply 4 times 72.

$$x=\frac{-\left(-1\right)±\sqrt{1+288}}{2\left(-1\right)}$$

Add 1 to 288.

$$x=\frac{-\left(-1\right)±\sqrt{289}}{2\left(-1\right)}$$

Take the square root of 289.

$$x=\frac{-\left(-1\right)±17}{2\left(-1\right)}$$

The opposite of -1 is 1.

$$x=\frac{1±17}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{1±17}{-2}$$

Now solve the equation x=\frac{1±17}{-2} when ± is plus. Add 1 to 17.

$$x=\frac{18}{-2}$$

Divide 18 by -2.

$$x=-9$$

Now solve the equation x=\frac{1±17}{-2} when ± is minus. Subtract 17 from 1.

$$x=-\frac{16}{-2}$$

Divide -16 by -2.

$$x=8$$

The equation is now solved.

$$x=-9 x=8$$

Steps for Completing the Square

Subtract x^{2} from both sides.

$$5x+72-x^{2}=6x$$

Subtract 6x from both sides.

$$5x+72-x^{2}-6x=0$$

Combine 5x and -6x to get -x.

$$-x+72-x^{2}=0$$

Subtract 72 from both sides. Anything subtracted from zero gives its negation.

$$-x-x^{2}=-72$$

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.

$$-x^{2}-x=-72$$

Divide both sides by -1.

$$\frac{-x^{2}-x}{-1}=-\frac{72}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{1}{-1}\right)x=-\frac{72}{-1}$$

Divide -1 by -1.

$$x^{2}+x=-\frac{72}{-1}$$

Divide -72 by -1.

$$x^{2}+x=72$$

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.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=72+\left(\frac{1}{2}\right)^{2}$$

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

$$x^{2}+x+\frac{1}{4}=72+\frac{1}{4}$$

Add 72 to \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=\frac{289}{4}$$

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

$$\left(x+\frac{1}{2}\right)^{2}=\frac{289}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{289}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{17}{2} x+\frac{1}{2}=-\frac{17}{2}$$

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

$$x=8 x=-9$$