Solve x^2-6+x^2+10x+18=0

$$x ^ { 2 } - 6 + x ^ { 2 } + 10 x + 18 = 0$$

Solve for x

$$x=-3$$

$$x=-2$$

Steps Using Factoring By Grouping

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

$$2x^{2}-6+10x+18=0$$

Add -6 and 18 to get 12.

$$2x^{2}+12+10x=0$$

Divide both sides by 2.

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

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

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.

$$a+b=5 ab=1\times 6=6$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 6.

$$1,6 2,3$$

Calculate the sum for each pair.

$$1+6=7 2+3=5$$

The solution is the pair that gives sum 5.

$$a=2 b=3$$

Rewrite x^{2}+5x+6 as \left(x^{2}+2x\right)+\left(3x+6\right).

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

Factor out x in the first and 3 in the second group.

$$x\left(x+2\right)+3\left(x+2\right)$$

Factor out common term x+2 by using distributive property.

$$\left(x+2\right)\left(x+3\right)$$

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

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

Steps Using the Quadratic Formula

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

$$2x^{2}-6+10x+18=0$$

Add -6 and 18 to get 12.

$$2x^{2}+12+10x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$2x^{2}+10x+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 10 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\times 2\times 12}}{2\times 2}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\times 2\times 12}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-10±\sqrt{100-8\times 12}}{2\times 2}$$

Multiply -8 times 12.

$$x=\frac{-10±\sqrt{100-96}}{2\times 2}$$

Add 100 to -96.

$$x=\frac{-10±\sqrt{4}}{2\times 2}$$

Take the square root of 4.

$$x=\frac{-10±2}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-10±2}{4}$$

Now solve the equation x=\frac{-10±2}{4} when ± is plus. Add -10 to 2.

$$x=-\frac{8}{4}$$

Divide -8 by 4.

$$x=-2$$

Now solve the equation x=\frac{-10±2}{4} when ± is minus. Subtract 2 from -10.

$$x=-\frac{12}{4}$$

Divide -12 by 4.

$$x=-3$$

The equation is now solved.

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

Steps for Completing the Square

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

$$2x^{2}-6+10x+18=0$$

Add -6 and 18 to get 12.

$$2x^{2}+12+10x=0$$

Subtract 12 from both sides. Anything subtracted from zero gives its negation.

$$2x^{2}+10x=-12$$

Divide both sides by 2.

$$\frac{2x^{2}+10x}{2}=-\frac{12}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{10}{2}x=-\frac{12}{2}$$

Divide 10 by 2.

$$x^{2}+5x=-\frac{12}{2}$$

Divide -12 by 2.

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

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}.

$$x^{2}+5x+\frac{25}{4}=\frac{1}{4}$$

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract \frac{5}{2} from both sides of the equation.

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