Solve (2)2x^2+6x+2=0

$$( 2 ) 2 x ^ { 2 } + 6 x + 2 = 0$$

Solve for x

$$x=-\frac{1}{2}=-0.5$$

$$x=-1$$

Steps Using Factoring By Grouping

Multiply 2 and 2 to get 4.

$$4x^{2}+6x+2=0$$

Divide both sides by 2.

$$2x^{2}+3x+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=3 ab=2\times 1=2$$

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.

$$a=1 b=2$$

Rewrite 2x^{2}+3x+1 as \left(2x^{2}+x\right)+\left(2x+1\right).

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

Factor out x in 2x^{2}+x.

$$x\left(2x+1\right)+2x+1$$

Factor out common term 2x+1 by using distributive property.

$$\left(2x+1\right)\left(x+1\right)$$

To find equation solutions, solve 2x+1=0 and x+1=0.

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

Steps Using the Quadratic Formula

Multiply 2 and 2 to get 4.

$$4x^{2}+6x+2=0$$

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

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

Square 6.

$$x=\frac{-6±\sqrt{36-4\times 4\times 2}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-6±\sqrt{36-16\times 2}}{2\times 4}$$

Multiply -16 times 2.

$$x=\frac{-6±\sqrt{36-32}}{2\times 4}$$

Add 36 to -32.

$$x=\frac{-6±\sqrt{4}}{2\times 4}$$

Take the square root of 4.

$$x=\frac{-6±2}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-6±2}{8}$$

Now solve the equation x=\frac{-6±2}{8} when ± is plus. Add -6 to 2.

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

Reduce the fraction \frac{-4}{8} to lowest terms by extracting and canceling out 4.

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

Now solve the equation x=\frac{-6±2}{8} when ± is minus. Subtract 2 from -6.

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

Divide -8 by 8.

$$x=-1$$

The equation is now solved.

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

Steps for Completing the Square

Multiply 2 and 2 to get 4.

$$4x^{2}+6x+2=0$$

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

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

Reduce the fraction \frac{6}{4} to lowest terms by extracting and canceling out 2.

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

Reduce the fraction \frac{-2}{4} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{3}{2}x=-\frac{1}{2}$$

Divide \frac{3}{2}, the coefficient of the x term, by 2 to get \frac{3}{4}. Then add the square of \frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{3}{2}x+\left(\frac{3}{4}\right)^{2}=-\frac{1}{2}+\left(\frac{3}{4}\right)^{2}$$

Square \frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=-\frac{1}{2}+\frac{9}{16}$$

Add -\frac{1}{2} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{1}{16}$$

Factor x^{2}+\frac{3}{2}x+\frac{9}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{4}\right)^{2}=\frac{1}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Simplify.

$$x+\frac{3}{4}=\frac{1}{4} x+\frac{3}{4}=-\frac{1}{4}$$

Subtract \frac{3}{4} from both sides of the equation.

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