Solve x^2-x+12=2x^2+3x+7

$$x ^ { 2 } - x + 12 = 2 x ^ { 2 } + 3 x + 7$$

Solve for x

$$x=-5$$

$$x=1$$

Steps Using Factoring By Grouping

Subtract 2x^{2} from both sides.

$$x^{2}-x+12-2x^{2}=3x+7$$

Combine x^{2} and -2x^{2} to get -x^{2}.

$$-x^{2}-x+12=3x+7$$

Subtract 3x from both sides.

$$-x^{2}-x+12-3x=7$$

Combine -x and -3x to get -4x.

$$-x^{2}-4x+12=7$$

Subtract 7 from both sides.

$$-x^{2}-4x+12-7=0$$

Subtract 7 from 12 to get 5.

$$-x^{2}-4x+5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+5. To find a and b, set up a system to be solved.

$$a+b=-4 ab=-5=-5$$

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. The only such pair is the system solution.

$$a=1 b=-5$$

Rewrite -x^{2}-4x+5 as \left(-x^{2}+x\right)+\left(-5x+5\right).

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

Factor out x in the first and 5 in the second group.

$$x\left(-x+1\right)+5\left(-x+1\right)$$

Factor out common term -x+1 by using distributive property.

$$\left(-x+1\right)\left(x+5\right)$$

To find equation solutions, solve -x+1=0 and x+5=0.

$$x=1 x=-5$$

Steps Using the Quadratic Formula

Subtract 2x^{2} from both sides.

$$x^{2}-x+12-2x^{2}=3x+7$$

Combine x^{2} and -2x^{2} to get -x^{2}.

$$-x^{2}-x+12=3x+7$$

Subtract 3x from both sides.

$$-x^{2}-x+12-3x=7$$

Combine -x and -3x to get -4x.

$$-x^{2}-4x+12=7$$

Subtract 7 from both sides.

$$-x^{2}-4x+12-7=0$$

Subtract 7 from 12 to get 5.

$$-x^{2}-4x+5=0$$

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

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-1\right)\times 5}}{2\left(-1\right)}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-1\right)\times 5}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-4\right)±\sqrt{16+4\times 5}}{2\left(-1\right)}$$

Multiply 4 times 5.

$$x=\frac{-\left(-4\right)±\sqrt{16+20}}{2\left(-1\right)}$$

Add 16 to 20.

$$x=\frac{-\left(-4\right)±\sqrt{36}}{2\left(-1\right)}$$

Take the square root of 36.

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

The opposite of -4 is 4.

$$x=\frac{4±6}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{4±6}{-2}$$

Now solve the equation x=\frac{4±6}{-2} when ± is plus. Add 4 to 6.

$$x=\frac{10}{-2}$$

Divide 10 by -2.

$$x=-5$$

Now solve the equation x=\frac{4±6}{-2} when ± is minus. Subtract 6 from 4.

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

Divide -2 by -2.

$$x=1$$

The equation is now solved.

$$x=-5 x=1$$

Steps for Completing the Square

Subtract 2x^{2} from both sides.

$$x^{2}-x+12-2x^{2}=3x+7$$

Combine x^{2} and -2x^{2} to get -x^{2}.

$$-x^{2}-x+12=3x+7$$

Subtract 3x from both sides.

$$-x^{2}-x+12-3x=7$$

Combine -x and -3x to get -4x.

$$-x^{2}-4x+12=7$$

Subtract 12 from both sides.

$$-x^{2}-4x=7-12$$

Subtract 12 from 7 to get -5.

$$-x^{2}-4x=-5$$

Divide both sides by -1.

$$\frac{-x^{2}-4x}{-1}=-\frac{5}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{4}{-1}\right)x=-\frac{5}{-1}$$

Divide -4 by -1.

$$x^{2}+4x=-\frac{5}{-1}$$

Divide -5 by -1.

$$x^{2}+4x=5$$

Divide 4, the coefficient of the x term, by 2 to get 2. Then add the square of 2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+4x+2^{2}=5+2^{2}$$

Square 2.

$$x^{2}+4x+4=5+4$$

Add 5 to 4.

$$x^{2}+4x+4=9$$

Factor x^{2}+4x+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}.

$$\left(x+2\right)^{2}=9$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{9}$$

Simplify.

$$x+2=3 x+2=-3$$

Subtract 2 from both sides of the equation.

$$x=1 x=-5$$