Solve x^2+20x+72=3x

$$x ^ { 2 } + 20 x + 72 = 3 x$$

Solve for x

$$x=-9$$

$$x=-8$$

Steps Using Factoring

Subtract 3x from both sides.

$$x^{2}+20x+72-3x=0$$

Combine 20x and -3x to get 17x.

$$x^{2}+17x+72=0$$

To solve the equation, factor x^{2}+17x+72 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=17 ab=72$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both 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.

$$1+72=73 2+36=38 3+24=27 4+18=22 6+12=18 8+9=17$$

The solution is the pair that gives sum 17.

$$a=8 b=9$$

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

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

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

$$x=-8 x=-9$$

Steps Using Factoring By Grouping

Subtract 3x from both sides.

$$x^{2}+20x+72-3x=0$$

Combine 20x and -3x to get 17x.

$$x^{2}+17x+72=0$$

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.

$$a+b=17 ab=1\times 72=72$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both 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.

$$1+72=73 2+36=38 3+24=27 4+18=22 6+12=18 8+9=17$$

The solution is the pair that gives sum 17.

$$a=8 b=9$$

Rewrite x^{2}+17x+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 3x from both sides.

$$x^{2}+20x+72-3x=0$$

Combine 20x and -3x to get 17x.

$$x^{2}+17x+72=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 17 for b, and 72 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-17±\sqrt{17^{2}-4\times 72}}{2}$$

Square 17.

$$x=\frac{-17±\sqrt{289-4\times 72}}{2}$$

Multiply -4 times 72.

$$x=\frac{-17±\sqrt{289-288}}{2}$$

Add 289 to -288.

$$x=\frac{-17±\sqrt{1}}{2}$$

Take the square root of 1.

$$x=\frac{-17±1}{2}$$

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

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

Divide -16 by 2.

$$x=-8$$

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

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

Divide -18 by 2.

$$x=-9$$

The equation is now solved.

$$x=-8 x=-9$$

Steps for Completing the Square

Subtract 3x from both sides.

$$x^{2}+20x+72-3x=0$$

Combine 20x and -3x to get 17x.

$$x^{2}+17x+72=0$$

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

$$x^{2}+17x=-72$$

Divide 17, the coefficient of the x term, by 2 to get \frac{17}{2}. Then add the square of \frac{17}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+17x+\left(\frac{17}{2}\right)^{2}=-72+\left(\frac{17}{2}\right)^{2}$$

Square \frac{17}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+17x+\frac{289}{4}=-72+\frac{289}{4}$$

Add -72 to \frac{289}{4}.

$$x^{2}+17x+\frac{289}{4}=\frac{1}{4}$$

Factor x^{2}+17x+\frac{289}{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}.

$$\left(x+\frac{17}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{17}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

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

Subtract \frac{17}{2} from both sides of the equation.

$$x=-8 x=-9$$