Solve 13x^2-26x-182=13

$$13 x ^ { 2 } - 26 x - 182 = 13$$

Solve for x

$$x=-3$$

$$x=5$$

Steps Using Factoring By Grouping

Subtract 13 from both sides.

$$13x^{2}-26x-182-13=0$$

Subtract 13 from -182 to get -195.

$$13x^{2}-26x-195=0$$

Divide both sides by 13.

$$x^{2}-2x-15=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-15. To find a and b, set up a system to be solved.

$$a+b=-2 ab=1\left(-15\right)=-15$$

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 -15.

$$1,-15 3,-5$$

Calculate the sum for each pair.

$$1-15=-14 3-5=-2$$

The solution is the pair that gives sum -2.

$$a=-5 b=3$$

Rewrite x^{2}-2x-15 as \left(x^{2}-5x\right)+\left(3x-15\right).

$$\left(x^{2}-5x\right)+\left(3x-15\right)$$

Factor out x in the first and 3 in the second group.

$$x\left(x-5\right)+3\left(x-5\right)$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(x+3\right)$$

To find equation solutions, solve x-5=0 and x+3=0.

$$x=5 x=-3$$

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.

$$13x^{2}-26x-182=13$$

Subtract 13 from both sides of the equation.

$$13x^{2}-26x-182-13=13-13$$

Subtracting 13 from itself leaves 0.

$$13x^{2}-26x-182-13=0$$

Subtract 13 from -182.

$$13x^{2}-26x-195=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 13 for a, -26 for b, and -195 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-26\right)±\sqrt{\left(-26\right)^{2}-4\times 13\left(-195\right)}}{2\times 13}$$

Square -26.

$$x=\frac{-\left(-26\right)±\sqrt{676-4\times 13\left(-195\right)}}{2\times 13}$$

Multiply -4 times 13.

$$x=\frac{-\left(-26\right)±\sqrt{676-52\left(-195\right)}}{2\times 13}$$

Multiply -52 times -195.

$$x=\frac{-\left(-26\right)±\sqrt{676+10140}}{2\times 13}$$

Add 676 to 10140.

$$x=\frac{-\left(-26\right)±\sqrt{10816}}{2\times 13}$$

Take the square root of 10816.

$$x=\frac{-\left(-26\right)±104}{2\times 13}$$

The opposite of -26 is 26.

$$x=\frac{26±104}{2\times 13}$$

Multiply 2 times 13.

$$x=\frac{26±104}{26}$$

Now solve the equation x=\frac{26±104}{26} when ± is plus. Add 26 to 104.

$$x=\frac{130}{26}$$

Divide 130 by 26.

$$x=5$$

Now solve the equation x=\frac{26±104}{26} when ± is minus. Subtract 104 from 26.

$$x=-\frac{78}{26}$$

Divide -78 by 26.

$$x=-3$$

The equation is now solved.

$$x=5 x=-3$$

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.

$$13x^{2}-26x-182=13$$

Add 182 to both sides of the equation.

$$13x^{2}-26x-182-\left(-182\right)=13-\left(-182\right)$$

Subtracting -182 from itself leaves 0.

$$13x^{2}-26x=13-\left(-182\right)$$

Subtract -182 from 13.

$$13x^{2}-26x=195$$

Divide both sides by 13.

$$\frac{13x^{2}-26x}{13}=\frac{195}{13}$$

Dividing by 13 undoes the multiplication by 13.

$$x^{2}+\left(-\frac{26}{13}\right)x=\frac{195}{13}$$

Divide -26 by 13.

$$x^{2}-2x=\frac{195}{13}$$

Divide 195 by 13.

$$x^{2}-2x=15$$

Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-2x+1=15+1$$

Add 15 to 1.

$$x^{2}-2x+1=16$$

Factor x^{2}-2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-1\right)^{2}=16$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{16}$$

Simplify.

$$x-1=4 x-1=-4$$

Add 1 to both sides of the equation.

$$x=5 x=-3$$