Solve 0=2(21+7x+6x+2x^2)

$$0 = 2 ( 21 + 7 x + 6 x + 2 x ^ { 2 } )$$

Solve for x

$$x = -\frac{7}{2} = -3\frac{1}{2} = -3.5$$

$$x=-3$$

Steps Using Factoring By Grouping

Divide both sides by 2. Zero divided by any non-zero number gives zero.

$$0=21+7x+6x+2x^{2}$$

Combine 7x and 6x to get 13x.

$$0=21+13x+2x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$21+13x+2x^{2}=0$$

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

$$2x^{2}+13x+21=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx+21. To find a and b, set up a system to be solved.

$$a+b=13 ab=2\times 21=42$$

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 42.

$$1,42 2,21 3,14 6,7$$

Calculate the sum for each pair.

$$1+42=43 2+21=23 3+14=17 6+7=13$$

The solution is the pair that gives sum 13.

$$a=6 b=7$$

Rewrite 2x^{2}+13x+21 as \left(2x^{2}+6x\right)+\left(7x+21\right).

$$\left(2x^{2}+6x\right)+\left(7x+21\right)$$

Factor out 2x in the first and 7 in the second group.

$$2x\left(x+3\right)+7\left(x+3\right)$$

Factor out common term x+3 by using distributive property.

$$\left(x+3\right)\left(2x+7\right)$$

To find equation solutions, solve x+3=0 and 2x+7=0.

$$x=-3 x=-\frac{7}{2}$$

Steps Using the Quadratic Formula

Divide both sides by 2. Zero divided by any non-zero number gives zero.

$$0=21+7x+6x+2x^{2}$$

Combine 7x and 6x to get 13x.

$$0=21+13x+2x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$21+13x+2x^{2}=0$$

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.

$$2x^{2}+13x+21=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 13 for b, and 21 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-13±\sqrt{13^{2}-4\times 2\times 21}}{2\times 2}$$

Square 13.

$$x=\frac{-13±\sqrt{169-4\times 2\times 21}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-13±\sqrt{169-8\times 21}}{2\times 2}$$

Multiply -8 times 21.

$$x=\frac{-13±\sqrt{169-168}}{2\times 2}$$

Add 169 to -168.

$$x=\frac{-13±\sqrt{1}}{2\times 2}$$

Take the square root of 1.

$$x=\frac{-13±1}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-13±1}{4}$$

Now solve the equation x=\frac{-13±1}{4} when ± is plus. Add -13 to 1.

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

Divide -12 by 4.

$$x=-3$$

Now solve the equation x=\frac{-13±1}{4} when ± is minus. Subtract 1 from -13.

$$x=-\frac{14}{4}$$

Reduce the fraction \frac{-14}{4} to lowest terms by extracting and canceling out 2.

$$x=-\frac{7}{2}$$

The equation is now solved.

$$x=-3 x=-\frac{7}{2}$$

Steps for Completing the Square

Divide both sides by 2. Zero divided by any non-zero number gives zero.

$$0=21+7x+6x+2x^{2}$$

Combine 7x and 6x to get 13x.

$$0=21+13x+2x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$21+13x+2x^{2}=0$$

Subtract 21 from both sides. Anything subtracted from zero gives its negation.

$$13x+2x^{2}=-21$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$2x^{2}+13x=-21$$

Divide both sides by 2.

$$\frac{2x^{2}+13x}{2}=-\frac{21}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{13}{2}x=-\frac{21}{2}$$

Divide \frac{13}{2}, the coefficient of the x term, by 2 to get \frac{13}{4}. Then add the square of \frac{13}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{13}{2}x+\left(\frac{13}{4}\right)^{2}=-\frac{21}{2}+\left(\frac{13}{4}\right)^{2}$$

Square \frac{13}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{13}{2}x+\frac{169}{16}=-\frac{21}{2}+\frac{169}{16}$$

Add -\frac{21}{2} to \frac{169}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{13}{2}x+\frac{169}{16}=\frac{1}{16}$$

Factor x^{2}+\frac{13}{2}x+\frac{169}{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}.

$$\left(x+\frac{13}{4}\right)^{2}=\frac{1}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{13}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Simplify.

$$x+\frac{13}{4}=\frac{1}{4} x+\frac{13}{4}=-\frac{1}{4}$$

Subtract \frac{13}{4} from both sides of the equation.

$$x=-3 x=-\frac{7}{2}$$