Solve x^2+x-6x-6=0

$$x ^ { 2 } + x - 6 x - 6 = 0$$

Solve for x

$$x=-1$$

$$x=6$$

Steps Using Factoring

Combine x and -6x to get -5x.

$$x^{2}-5x-6=0$$

To solve the equation, factor x^{2}-5x-6 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.

$$a+b=-5 ab=-6$$

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 -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -5.

$$a=-6 b=1$$

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

$$\left(x-6\right)\left(x+1\right)$$

To find equation solutions, solve x-6=0 and x+1=0.

$$x=6 x=-1$$

Steps Using Factoring By Grouping

Combine x and -6x to get -5x.

$$x^{2}-5x-6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

$$a+b=-5 ab=1\left(-6\right)=-6$$

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 -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -5.

$$a=-6 b=1$$

Rewrite x^{2}-5x-6 as \left(x^{2}-6x\right)+\left(x-6\right).

$$\left(x^{2}-6x\right)+\left(x-6\right)$$

Factor out x in x^{2}-6x.

$$x\left(x-6\right)+x-6$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(x+1\right)$$

To find equation solutions, solve x-6=0 and x+1=0.

$$x=6 x=-1$$

Steps Using the Quadratic Formula

Combine x and -6x to get -5x.

$$x^{2}-5x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -5 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-6\right)}}{2}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-6\right)}}{2}$$

Multiply -4 times -6.

$$x=\frac{-\left(-5\right)±\sqrt{25+24}}{2}$$

Add 25 to 24.

$$x=\frac{-\left(-5\right)±\sqrt{49}}{2}$$

Take the square root of 49.

$$x=\frac{-\left(-5\right)±7}{2}$$

The opposite of -5 is 5.

$$x=\frac{5±7}{2}$$

Now solve the equation x=\frac{5±7}{2} when ± is plus. Add 5 to 7.

$$x=\frac{12}{2}$$

Divide 12 by 2.

$$x=6$$

Now solve the equation x=\frac{5±7}{2} when ± is minus. Subtract 7 from 5.

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

Divide -2 by 2.

$$x=-1$$

The equation is now solved.

$$x=6 x=-1$$

Steps for Completing the Square

Combine x and -6x to get -5x.

$$x^{2}-5x-6=0$$

Add 6 to both sides. Anything plus zero gives itself.

$$x^{2}-5x=6$$

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=6+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=6+\frac{25}{4}$$

Add 6 to \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{49}{4}$$

Factor x^{2}-5x+\frac{25}{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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{7}{2} x-\frac{5}{2}=-\frac{7}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$x=6 x=-1$$