Solve 180=3x+x^2

$$180 = 3 x + x ^ { 2 }$$

Solve for x

$$x=-15$$

$$x=12$$

Steps Using Factoring

Swap sides so that all variable terms are on the left hand side.

$$3x+x^{2}=180$$

Subtract 180 from both sides.

$$3x+x^{2}-180=0$$

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

$$x^{2}+3x-180=0$$

To solve the equation, factor x^{2}+3x-180 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.

$$a+b=3 ab=-180$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -180.

$$-1,180 -2,90 -3,60 -4,45 -5,36 -6,30 -9,20 -10,18 -12,15$$

Calculate the sum for each pair.

$$-1+180=179 -2+90=88 -3+60=57 -4+45=41 -5+36=31 -6+30=24 -9+20=11 -10+18=8 -12+15=3$$

The solution is the pair that gives sum 3.

$$a=-12 b=15$$

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

$$\left(x-12\right)\left(x+15\right)$$

To find equation solutions, solve x-12=0 and x+15=0.

$$x=12 x=-15$$

Steps Using Factoring By Grouping

Swap sides so that all variable terms are on the left hand side.

$$3x+x^{2}=180$$

Subtract 180 from both sides.

$$3x+x^{2}-180=0$$

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

$$x^{2}+3x-180=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-180. To find a and b, set up a system to be solved.

$$a+b=3 ab=1\left(-180\right)=-180$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -180.

$$-1,180 -2,90 -3,60 -4,45 -5,36 -6,30 -9,20 -10,18 -12,15$$

Calculate the sum for each pair.

$$-1+180=179 -2+90=88 -3+60=57 -4+45=41 -5+36=31 -6+30=24 -9+20=11 -10+18=8 -12+15=3$$

The solution is the pair that gives sum 3.

$$a=-12 b=15$$

Rewrite x^{2}+3x-180 as \left(x^{2}-12x\right)+\left(15x-180\right).

$$\left(x^{2}-12x\right)+\left(15x-180\right)$$

Factor out x in the first and 15 in the second group.

$$x\left(x-12\right)+15\left(x-12\right)$$

Factor out common term x-12 by using distributive property.

$$\left(x-12\right)\left(x+15\right)$$

To find equation solutions, solve x-12=0 and x+15=0.

$$x=12 x=-15$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$3x+x^{2}=180$$

Subtract 180 from both sides.

$$3x+x^{2}-180=0$$

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.

$$x^{2}+3x-180=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 3 for b, and -180 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-180\right)}}{2}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-180\right)}}{2}$$

Multiply -4 times -180.

$$x=\frac{-3±\sqrt{9+720}}{2}$$

Add 9 to 720.

$$x=\frac{-3±\sqrt{729}}{2}$$

Take the square root of 729.

$$x=\frac{-3±27}{2}$$

Now solve the equation x=\frac{-3±27}{2} when ± is plus. Add -3 to 27.

$$x=\frac{24}{2}$$

Divide 24 by 2.

$$x=12$$

Now solve the equation x=\frac{-3±27}{2} when ± is minus. Subtract 27 from -3.

$$x=-\frac{30}{2}$$

Divide -30 by 2.

$$x=-15$$

The equation is now solved.

$$x=12 x=-15$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$3x+x^{2}=180$$

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.

$$x^{2}+3x=180$$

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.

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=180+\left(\frac{3}{2}\right)^{2}$$

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

$$x^{2}+3x+\frac{9}{4}=180+\frac{9}{4}$$

Add 180 to \frac{9}{4}.

$$x^{2}+3x+\frac{9}{4}=\frac{729}{4}$$

Factor x^{2}+3x+\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}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{729}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{729}{4}}$$

Simplify.

$$x+\frac{3}{2}=\frac{27}{2} x+\frac{3}{2}=-\frac{27}{2}$$

Subtract \frac{3}{2} from both sides of the equation.

$$x=12 x=-15$$