Solve x^2+30x-110=1265

$$x ^ { 2 } + 30 x - 110 = 1265$$

Solve for x

$$x=-55$$

$$x=25$$

Steps Using Factoring

Subtract 1265 from both sides.

$$x^{2}+30x-110-1265=0$$

Subtract 1265 from -110 to get -1375.

$$x^{2}+30x-1375=0$$

To solve the equation, factor x^{2}+30x-1375 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.

$$a+b=30 ab=-1375$$

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 -1375.

$$-1,1375 -5,275 -11,125 -25,55$$

Calculate the sum for each pair.

$$-1+1375=1374 -5+275=270 -11+125=114 -25+55=30$$

The solution is the pair that gives sum 30.

$$a=-25 b=55$$

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

$$\left(x-25\right)\left(x+55\right)$$

To find equation solutions, solve x-25=0 and x+55=0.

$$x=25 x=-55$$

Steps Using Factoring By Grouping

Subtract 1265 from both sides.

$$x^{2}+30x-110-1265=0$$

Subtract 1265 from -110 to get -1375.

$$x^{2}+30x-1375=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-1375. To find a and b, set up a system to be solved.

$$a+b=30 ab=1\left(-1375\right)=-1375$$

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 -1375.

$$-1,1375 -5,275 -11,125 -25,55$$

Calculate the sum for each pair.

$$-1+1375=1374 -5+275=270 -11+125=114 -25+55=30$$

The solution is the pair that gives sum 30.

$$a=-25 b=55$$

Rewrite x^{2}+30x-1375 as \left(x^{2}-25x\right)+\left(55x-1375\right).

$$\left(x^{2}-25x\right)+\left(55x-1375\right)$$

Factor out x in the first and 55 in the second group.

$$x\left(x-25\right)+55\left(x-25\right)$$

Factor out common term x-25 by using distributive property.

$$\left(x-25\right)\left(x+55\right)$$

To find equation solutions, solve x-25=0 and x+55=0.

$$x=25 x=-55$$

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.

$$x^{2}+30x-110=1265$$

Subtract 1265 from both sides of the equation.

$$x^{2}+30x-110-1265=1265-1265$$

Subtracting 1265 from itself leaves 0.

$$x^{2}+30x-110-1265=0$$

Subtract 1265 from -110.

$$x^{2}+30x-1375=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 30 for b, and -1375 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-30±\sqrt{30^{2}-4\left(-1375\right)}}{2}$$

Square 30.

$$x=\frac{-30±\sqrt{900-4\left(-1375\right)}}{2}$$

Multiply -4 times -1375.

$$x=\frac{-30±\sqrt{900+5500}}{2}$$

Add 900 to 5500.

$$x=\frac{-30±\sqrt{6400}}{2}$$

Take the square root of 6400.

$$x=\frac{-30±80}{2}$$

Now solve the equation x=\frac{-30±80}{2} when ± is plus. Add -30 to 80.

$$x=\frac{50}{2}$$

Divide 50 by 2.

$$x=25$$

Now solve the equation x=\frac{-30±80}{2} when ± is minus. Subtract 80 from -30.

$$x=-\frac{110}{2}$$

Divide -110 by 2.

$$x=-55$$

The equation is now solved.

$$x=25 x=-55$$

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.

$$x^{2}+30x-110=1265$$

Add 110 to both sides of the equation.

$$x^{2}+30x-110-\left(-110\right)=1265-\left(-110\right)$$

Subtracting -110 from itself leaves 0.

$$x^{2}+30x=1265-\left(-110\right)$$

Subtract -110 from 1265.

$$x^{2}+30x=1375$$

Divide 30, the coefficient of the x term, by 2 to get 15. Then add the square of 15 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+30x+15^{2}=1375+15^{2}$$

Square 15.

$$x^{2}+30x+225=1375+225$$

Add 1375 to 225.

$$x^{2}+30x+225=1600$$

Factor x^{2}+30x+225. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+15\right)^{2}=1600$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+15\right)^{2}}=\sqrt{1600}$$

Simplify.

$$x+15=40 x+15=-40$$

Subtract 15 from both sides of the equation.

$$x=25 x=-55$$