Solve x^2+x^2+8x+6=0

$$x ^ { 2 } + x ^ { 2 } + 8 x + 6 = 0$$

Solve for x

$$x=-3$$

$$x=-1$$

Steps Using Factoring By Grouping

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

$$2x^{2}+8x+6=0$$

Divide both sides by 2.

$$x^{2}+4x+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=4 ab=1\times 3=3$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=3$$

Rewrite x^{2}+4x+3 as \left(x^{2}+x\right)+\left(3x+3\right).

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

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

$$x\left(x+1\right)+3\left(x+1\right)$$

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

$$\left(x+1\right)\left(x+3\right)$$

To find equation solutions, solve x+1=0 and x+3=0.

$$x=-1 x=-3$$

Steps Using the Quadratic Formula

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

$$2x^{2}+8x+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 8 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\times 2\times 6}}{2\times 2}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\times 2\times 6}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-8±\sqrt{64-8\times 6}}{2\times 2}$$

Multiply -8 times 6.

$$x=\frac{-8±\sqrt{64-48}}{2\times 2}$$

Add 64 to -48.

$$x=\frac{-8±\sqrt{16}}{2\times 2}$$

Take the square root of 16.

$$x=\frac{-8±4}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-8±4}{4}$$

Now solve the equation x=\frac{-8±4}{4} when ± is plus. Add -8 to 4.

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

Divide -4 by 4.

$$x=-1$$

Now solve the equation x=\frac{-8±4}{4} when ± is minus. Subtract 4 from -8.

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

Divide -12 by 4.

$$x=-3$$

The equation is now solved.

$$x=-1 x=-3$$

Steps for Completing the Square

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

$$2x^{2}+8x+6=0$$

Subtract 6 from both sides. Anything subtracted from zero gives its negation.

$$2x^{2}+8x=-6$$

Divide both sides by 2.

$$\frac{2x^{2}+8x}{2}=-\frac{6}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{8}{2}x=-\frac{6}{2}$$

Divide 8 by 2.

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

Divide -6 by 2.

$$x^{2}+4x=-3$$

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.

$$x^{2}+4x+2^{2}=-3+2^{2}$$

Square 2.

$$x^{2}+4x+4=-3+4$$

Add -3 to 4.

$$x^{2}+4x+4=1$$

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+2=1 x+2=-1$$

Subtract 2 from both sides of the equation.

$$x=-1 x=-3$$