Solve x^2+13x+35=x

$$x ^ { 2 } + 13 x + 35 = x$$

Solve for x

$$x=-7$$

$$x=-5$$

Steps Using Factoring

Subtract x from both sides.

$$x^{2}+13x+35-x=0$$

Combine 13x and -x to get 12x.

$$x^{2}+12x+35=0$$

To solve the equation, factor x^{2}+12x+35 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.

$$a+b=12 ab=35$$

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 35.

$$1,35 5,7$$

Calculate the sum for each pair.

$$1+35=36 5+7=12$$

The solution is the pair that gives sum 12.

$$a=5 b=7$$

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

$$\left(x+5\right)\left(x+7\right)$$

To find equation solutions, solve x+5=0 and x+7=0.

$$x=-5 x=-7$$

Steps Using Factoring By Grouping

Subtract x from both sides.

$$x^{2}+13x+35-x=0$$

Combine 13x and -x to get 12x.

$$x^{2}+12x+35=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+35. To find a and b, set up a system to be solved.

$$a+b=12 ab=1\times 35=35$$

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 35.

$$1,35 5,7$$

Calculate the sum for each pair.

$$1+35=36 5+7=12$$

The solution is the pair that gives sum 12.

$$a=5 b=7$$

Rewrite x^{2}+12x+35 as \left(x^{2}+5x\right)+\left(7x+35\right).

$$\left(x^{2}+5x\right)+\left(7x+35\right)$$

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

$$x\left(x+5\right)+7\left(x+5\right)$$

Factor out common term x+5 by using distributive property.

$$\left(x+5\right)\left(x+7\right)$$

To find equation solutions, solve x+5=0 and x+7=0.

$$x=-5 x=-7$$

Steps Using the Quadratic Formula

Subtract x from both sides.

$$x^{2}+13x+35-x=0$$

Combine 13x and -x to get 12x.

$$x^{2}+12x+35=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 12 for b, and 35 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\times 35}}{2}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\times 35}}{2}$$

Multiply -4 times 35.

$$x=\frac{-12±\sqrt{144-140}}{2}$$

Add 144 to -140.

$$x=\frac{-12±\sqrt{4}}{2}$$

Take the square root of 4.

$$x=\frac{-12±2}{2}$$

Now solve the equation x=\frac{-12±2}{2} when ± is plus. Add -12 to 2.

$$x=-\frac{10}{2}$$

Divide -10 by 2.

$$x=-5$$

Now solve the equation x=\frac{-12±2}{2} when ± is minus. Subtract 2 from -12.

$$x=-\frac{14}{2}$$

Divide -14 by 2.

$$x=-7$$

The equation is now solved.

$$x=-5 x=-7$$

Steps for Completing the Square

Subtract x from both sides.

$$x^{2}+13x+35-x=0$$

Combine 13x and -x to get 12x.

$$x^{2}+12x+35=0$$

Subtract 35 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}+12x=-35$$

Divide 12, the coefficient of the x term, by 2 to get 6. Then add the square of 6 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+12x+6^{2}=-35+6^{2}$$

Square 6.

$$x^{2}+12x+36=-35+36$$

Add -35 to 36.

$$x^{2}+12x+36=1$$

Factor x^{2}+12x+36. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+6\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+6\right)^{2}}=\sqrt{1}$$

Simplify.

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

Subtract 6 from both sides of the equation.

$$x=-5 x=-7$$