Solve X^2=X+6

$$X ^ { 2 } = X + 6$$

Solve for X

$$X=-2$$

$$X=3$$

Steps Using Factoring

Subtract X from both sides.

$$X^{2}-X=6$$

Subtract 6 from both sides.

$$X^{2}-X-6=0$$

To solve the equation, factor X^{2}-X-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.

$$a+b=-1 ab=-6$$

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 -1.

$$a=-3 b=2$$

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

$$\left(X-3\right)\left(X+2\right)$$

To find equation solutions, solve X-3=0 and X+2=0.

$$X=3 X=-2$$

Steps Using Factoring By Grouping

Subtract X from both sides.

$$X^{2}-X=6$$

Subtract 6 from both sides.

$$X^{2}-X-6=0$$

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.

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

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 -1.

$$a=-3 b=2$$

Rewrite X^{2}-X-6 as \left(X^{2}-3X\right)+\left(2X-6\right).

$$\left(X^{2}-3X\right)+\left(2X-6\right)$$

Factor out X in the first and 2 in the second group.

$$X\left(X-3\right)+2\left(X-3\right)$$

Factor out common term X-3 by using distributive property.

$$\left(X-3\right)\left(X+2\right)$$

To find equation solutions, solve X-3=0 and X+2=0.

$$X=3 X=-2$$

Steps Using the Quadratic Formula

Subtract X from both sides.

$$X^{2}-X=6$$

Subtract 6 from both sides.

$$X^{2}-X-6=0$$

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

$$X=\frac{-\left(-1\right)±\sqrt{1-4\left(-6\right)}}{2}$$

Multiply -4 times -6.

$$X=\frac{-\left(-1\right)±\sqrt{1+24}}{2}$$

Add 1 to 24.

$$X=\frac{-\left(-1\right)±\sqrt{25}}{2}$$

Take the square root of 25.

$$X=\frac{-\left(-1\right)±5}{2}$$

The opposite of -1 is 1.

$$X=\frac{1±5}{2}$$

Now solve the equation X=\frac{1±5}{2} when ± is plus. Add 1 to 5.

$$X=\frac{6}{2}$$

Divide 6 by 2.

$$X=3$$

Now solve the equation X=\frac{1±5}{2} when ± is minus. Subtract 5 from 1.

$$X=-\frac{4}{2}$$

Divide -4 by 2.

$$X=-2$$

The equation is now solved.

$$X=3 X=-2$$

Steps for Completing the Square

Subtract X from both sides.

$$X^{2}-X=6$$

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

$$X^{2}-X+\left(-\frac{1}{2}\right)^{2}=6+\left(-\frac{1}{2}\right)^{2}$$

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$X^{2}-X+\frac{1}{4}=6+\frac{1}{4}$$

Add 6 to \frac{1}{4}.

$$X^{2}-X+\frac{1}{4}=\frac{25}{4}$$

Factor X^{2}-X+\frac{1}{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}.

$$\left(X-\frac{1}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(X-\frac{1}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$X-\frac{1}{2}=\frac{5}{2} X-\frac{1}{2}=-\frac{5}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$X=3 X=-2$$