Solve x^2+26x-360=0

$${ x }^{ 2 } +26x-360=0$$

Solve for x

$$x=-36$$

$$x=10$$

Steps Using Factoring

To solve the equation, factor x^{2}+26x-360 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.

$$a+b=26 ab=-360$$

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 -360.

$$-1,360 -2,180 -3,120 -4,90 -5,72 -6,60 -8,45 -9,40 -10,36 -12,30 -15,24 -18,20$$

Calculate the sum for each pair.

$$-1+360=359 -2+180=178 -3+120=117 -4+90=86 -5+72=67 -6+60=54 -8+45=37 -9+40=31 -10+36=26 -12+30=18 -15+24=9 -18+20=2$$

The solution is the pair that gives sum 26.

$$a=-10 b=36$$

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

$$\left(x-10\right)\left(x+36\right)$$

To find equation solutions, solve x-10=0 and x+36=0.

$$x=10 x=-36$$

Steps Using Factoring By Grouping

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-360. To find a and b, set up a system to be solved.

$$a+b=26 ab=1\left(-360\right)=-360$$

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 -360.

$$-1,360 -2,180 -3,120 -4,90 -5,72 -6,60 -8,45 -9,40 -10,36 -12,30 -15,24 -18,20$$

Calculate the sum for each pair.

$$-1+360=359 -2+180=178 -3+120=117 -4+90=86 -5+72=67 -6+60=54 -8+45=37 -9+40=31 -10+36=26 -12+30=18 -15+24=9 -18+20=2$$

The solution is the pair that gives sum 26.

$$a=-10 b=36$$

Rewrite x^{2}+26x-360 as \left(x^{2}-10x\right)+\left(36x-360\right).

$$\left(x^{2}-10x\right)+\left(36x-360\right)$$

Factor out x in the first and 36 in the second group.

$$x\left(x-10\right)+36\left(x-10\right)$$

Factor out common term x-10 by using distributive property.

$$\left(x-10\right)\left(x+36\right)$$

To find equation solutions, solve x-10=0 and x+36=0.

$$x=10 x=-36$$

Steps Using the Quadratic Formula

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.

$$x^{2}+26x-360=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 26 for b, and -360 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-26±\sqrt{26^{2}-4\left(-360\right)}}{2}$$

Square 26.

$$x=\frac{-26±\sqrt{676-4\left(-360\right)}}{2}$$

Multiply -4 times -360.

$$x=\frac{-26±\sqrt{676+1440}}{2}$$

Add 676 to 1440.

$$x=\frac{-26±\sqrt{2116}}{2}$$

Take the square root of 2116.

$$x=\frac{-26±46}{2}$$

Now solve the equation x=\frac{-26±46}{2} when ± is plus. Add -26 to 46.

$$x=\frac{20}{2}$$

Divide 20 by 2.

$$x=10$$

Now solve the equation x=\frac{-26±46}{2} when ± is minus. Subtract 46 from -26.

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

Divide -72 by 2.

$$x=-36$$

The equation is now solved.

$$x=10 x=-36$$

Steps for Completing the Square

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.

$$x^{2}+26x-360=0$$

Add 360 to both sides of the equation.

$$x^{2}+26x-360-\left(-360\right)=-\left(-360\right)$$

Subtracting -360 from itself leaves 0.

$$x^{2}+26x=-\left(-360\right)$$

Subtract -360 from 0.

$$x^{2}+26x=360$$

Divide 26, the coefficient of the x term, by 2 to get 13. Then add the square of 13 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+26x+13^{2}=360+13^{2}$$

Square 13.

$$x^{2}+26x+169=360+169$$

Add 360 to 169.

$$x^{2}+26x+169=529$$

Factor x^{2}+26x+169. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+13\right)^{2}=529$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+13\right)^{2}}=\sqrt{529}$$

Simplify.

$$x+13=23 x+13=-23$$

Subtract 13 from both sides of the equation.

$$x=10 x=-36$$