Solve 0=x^2+7x+6

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

Solve for x

$$x=-6$$

$$x=-1$$

Steps Using Factoring

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

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

To solve the equation, factor x^{2}+7x+6 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.

$$a+b=7 ab=6$$

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 7.

$$a=1 b=6$$

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

$$\left(x+1\right)\left(x+6\right)$$

To find equation solutions, solve x+1=0 and x+6=0.

$$x=-1 x=-6$$

Steps Using Factoring By Grouping

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

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

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.

$$a+b=7 ab=1\times 6=6$$

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 7.

$$a=1 b=6$$

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

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

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

$$x\left(x+1\right)+6\left(x+1\right)$$

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

$$\left(x+1\right)\left(x+6\right)$$

To find equation solutions, solve x+1=0 and x+6=0.

$$x=-1 x=-6$$

Steps Using the Quadratic Formula

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

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 7 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\times 6}}{2}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\times 6}}{2}$$

Multiply -4 times 6.

$$x=\frac{-7±\sqrt{49-24}}{2}$$

Add 49 to -24.

$$x=\frac{-7±\sqrt{25}}{2}$$

Take the square root of 25.

$$x=\frac{-7±5}{2}$$

Now solve the equation x=\frac{-7±5}{2} when ± is plus. Add -7 to 5.

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

Divide -2 by 2.

$$x=-1$$

Now solve the equation x=\frac{-7±5}{2} when ± is minus. Subtract 5 from -7.

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

Divide -12 by 2.

$$x=-6$$

The equation is now solved.

$$x=-1 x=-6$$

Steps for Completing the Square

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

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

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

$$x^{2}+7x=-6$$

Divide 7, the coefficient of the x term, by 2 to get \frac{7}{2}. Then add the square of \frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+7x+\left(\frac{7}{2}\right)^{2}=-6+\left(\frac{7}{2}\right)^{2}$$

Square \frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+7x+\frac{49}{4}=-6+\frac{49}{4}$$

Add -6 to \frac{49}{4}.

$$x^{2}+7x+\frac{49}{4}=\frac{25}{4}$$

Factor x^{2}+7x+\frac{49}{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}.

$$\left(x+\frac{7}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$x+\frac{7}{2}=\frac{5}{2} x+\frac{7}{2}=-\frac{5}{2}$$

Subtract \frac{7}{2} from both sides of the equation.

$$x=-1 x=-6$$