Solve x^2-18x+7x-126=0

$${ x }^{ 2 } -18x+7x-126=0$$

Solve for x

$$x=-7$$

$$x=18$$

Steps Using Factoring

Combine -18x and 7x to get -11x.

$$x^{2}-11x-126=0$$

To solve the equation, factor x^{2}-11x-126 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.

$$a+b=-11 ab=-126$$

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 -126.

$$1,-126 2,-63 3,-42 6,-21 7,-18 9,-14$$

Calculate the sum for each pair.

$$1-126=-125 2-63=-61 3-42=-39 6-21=-15 7-18=-11 9-14=-5$$

The solution is the pair that gives sum -11.

$$a=-18 b=7$$

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

$$\left(x-18\right)\left(x+7\right)$$

To find equation solutions, solve x-18=0 and x+7=0.

$$x=18 x=-7$$

Steps Using Factoring By Grouping

Combine -18x and 7x to get -11x.

$$x^{2}-11x-126=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-126. To find a and b, set up a system to be solved.

$$a+b=-11 ab=1\left(-126\right)=-126$$

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 -126.

$$1,-126 2,-63 3,-42 6,-21 7,-18 9,-14$$

Calculate the sum for each pair.

$$1-126=-125 2-63=-61 3-42=-39 6-21=-15 7-18=-11 9-14=-5$$

The solution is the pair that gives sum -11.

$$a=-18 b=7$$

Rewrite x^{2}-11x-126 as \left(x^{2}-18x\right)+\left(7x-126\right).

$$\left(x^{2}-18x\right)+\left(7x-126\right)$$

Factor out x in the first and 7 in the second group.

$$x\left(x-18\right)+7\left(x-18\right)$$

Factor out common term x-18 by using distributive property.

$$\left(x-18\right)\left(x+7\right)$$

To find equation solutions, solve x-18=0 and x+7=0.

$$x=18 x=-7$$

Steps Using the Quadratic Formula

Combine -18x and 7x to get -11x.

$$x^{2}-11x-126=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -11 for b, and -126 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\left(-126\right)}}{2}$$

Square -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\left(-126\right)}}{2}$$

Multiply -4 times -126.

$$x=\frac{-\left(-11\right)±\sqrt{121+504}}{2}$$

Add 121 to 504.

$$x=\frac{-\left(-11\right)±\sqrt{625}}{2}$$

Take the square root of 625.

$$x=\frac{-\left(-11\right)±25}{2}$$

The opposite of -11 is 11.

$$x=\frac{11±25}{2}$$

Now solve the equation x=\frac{11±25}{2} when ± is plus. Add 11 to 25.

$$x=\frac{36}{2}$$

Divide 36 by 2.

$$x=18$$

Now solve the equation x=\frac{11±25}{2} when ± is minus. Subtract 25 from 11.

$$x=-\frac{14}{2}$$

Divide -14 by 2.

$$x=-7$$

The equation is now solved.

$$x=18 x=-7$$

Steps for Completing the Square

Combine -18x and 7x to get -11x.

$$x^{2}-11x-126=0$$

Add 126 to both sides. Anything plus zero gives itself.

$$x^{2}-11x=126$$

Divide -11, the coefficient of the x term, by 2 to get -\frac{11}{2}. Then add the square of -\frac{11}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-11x+\left(-\frac{11}{2}\right)^{2}=126+\left(-\frac{11}{2}\right)^{2}$$

Square -\frac{11}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-11x+\frac{121}{4}=126+\frac{121}{4}$$

Add 126 to \frac{121}{4}.

$$x^{2}-11x+\frac{121}{4}=\frac{625}{4}$$

Factor x^{2}-11x+\frac{121}{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}.

$$\left(x-\frac{11}{2}\right)^{2}=\frac{625}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{2}\right)^{2}}=\sqrt{\frac{625}{4}}$$

Simplify.

$$x-\frac{11}{2}=\frac{25}{2} x-\frac{11}{2}=-\frac{25}{2}$$

Add \frac{11}{2} to both sides of the equation.

$$x=18 x=-7$$