Solve 2(x^2+x-8)=x^2+x+14

$$2( { x }^{ 2 } +x-8)= { x }^{ 2 } +x+14$$

Solve for x

$$x=-6$$

$$x=5$$

Steps Using Factoring

Use the distributive property to multiply 2 by x^{2}+x-8.

$$2x^{2}+2x-16=x^{2}+x+14$$

Subtract x^{2} from both sides.

$$2x^{2}+2x-16-x^{2}=x+14$$

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

$$x^{2}+2x-16=x+14$$

Subtract x from both sides.

$$x^{2}+2x-16-x=14$$

Combine 2x and -x to get x.

$$x^{2}+x-16=14$$

Subtract 14 from both sides.

$$x^{2}+x-16-14=0$$

Subtract 14 from -16 to get -30.

$$x^{2}+x-30=0$$

To solve the equation, factor x^{2}+x-30 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.

$$a+b=1 ab=-30$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -30.

$$-1,30 -2,15 -3,10 -5,6$$

Calculate the sum for each pair.

$$-1+30=29 -2+15=13 -3+10=7 -5+6=1$$

The solution is the pair that gives sum 1.

$$a=-5 b=6$$

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

$$\left(x-5\right)\left(x+6\right)$$

To find equation solutions, solve x-5=0 and x+6=0.

$$x=5 x=-6$$

Steps Using Factoring By Grouping

Use the distributive property to multiply 2 by x^{2}+x-8.

$$2x^{2}+2x-16=x^{2}+x+14$$

Subtract x^{2} from both sides.

$$2x^{2}+2x-16-x^{2}=x+14$$

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

$$x^{2}+2x-16=x+14$$

Subtract x from both sides.

$$x^{2}+2x-16-x=14$$

Combine 2x and -x to get x.

$$x^{2}+x-16=14$$

Subtract 14 from both sides.

$$x^{2}+x-16-14=0$$

Subtract 14 from -16 to get -30.

$$x^{2}+x-30=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-30. To find a and b, set up a system to be solved.

$$a+b=1 ab=1\left(-30\right)=-30$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -30.

$$-1,30 -2,15 -3,10 -5,6$$

Calculate the sum for each pair.

$$-1+30=29 -2+15=13 -3+10=7 -5+6=1$$

The solution is the pair that gives sum 1.

$$a=-5 b=6$$

Rewrite x^{2}+x-30 as \left(x^{2}-5x\right)+\left(6x-30\right).

$$\left(x^{2}-5x\right)+\left(6x-30\right)$$

Factor out x in the first and 6 in the second group.

$$x\left(x-5\right)+6\left(x-5\right)$$

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

$$\left(x-5\right)\left(x+6\right)$$

To find equation solutions, solve x-5=0 and x+6=0.

$$x=5 x=-6$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2 by x^{2}+x-8.

$$2x^{2}+2x-16=x^{2}+x+14$$

Subtract x^{2} from both sides.

$$2x^{2}+2x-16-x^{2}=x+14$$

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

$$x^{2}+2x-16=x+14$$

Subtract x from both sides.

$$x^{2}+2x-16-x=14$$

Combine 2x and -x to get x.

$$x^{2}+x-16=14$$

Subtract 14 from both sides.

$$x^{2}+x-16-14=0$$

Subtract 14 from -16 to get -30.

$$x^{2}+x-30=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 1 for b, and -30 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\left(-30\right)}}{2}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\left(-30\right)}}{2}$$

Multiply -4 times -30.

$$x=\frac{-1±\sqrt{1+120}}{2}$$

Add 1 to 120.

$$x=\frac{-1±\sqrt{121}}{2}$$

Take the square root of 121.

$$x=\frac{-1±11}{2}$$

Now solve the equation x=\frac{-1±11}{2} when ± is plus. Add -1 to 11.

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

Divide 10 by 2.

$$x=5$$

Now solve the equation x=\frac{-1±11}{2} when ± is minus. Subtract 11 from -1.

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

Divide -12 by 2.

$$x=-6$$

The equation is now solved.

$$x=5 x=-6$$

Steps for Completing the Square

Use the distributive property to multiply 2 by x^{2}+x-8.

$$2x^{2}+2x-16=x^{2}+x+14$$

Subtract x^{2} from both sides.

$$2x^{2}+2x-16-x^{2}=x+14$$

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

$$x^{2}+2x-16=x+14$$

Subtract x from both sides.

$$x^{2}+2x-16-x=14$$

Combine 2x and -x to get x.

$$x^{2}+x-16=14$$

Add 16 to both sides.

$$x^{2}+x=14+16$$

Add 14 and 16 to get 30.

$$x^{2}+x=30$$

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.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=30+\left(\frac{1}{2}\right)^{2}$$

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

$$x^{2}+x+\frac{1}{4}=30+\frac{1}{4}$$

Add 30 to \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=\frac{121}{4}$$

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}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{121}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{121}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{11}{2} x+\frac{1}{2}=-\frac{11}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$x=5 x=-6$$