Solve x^2+6=-5x

$$x ^ { 2 } + 6 = - 5 x$$

Solve for x

$$x=-3$$

$$x=-2$$

Steps Using Factoring

Add 5x to both sides.

$$x^{2}+6+5x=0$$

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

$$x^{2}+5x+6=0$$

To solve the equation, factor x^{2}+5x+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.

$$a+b=5 ab=6$$

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 5.

$$a=2 b=3$$

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

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

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

$$x=-2 x=-3$$

Steps Using Factoring By Grouping

Add 5x to both sides.

$$x^{2}+6+5x=0$$

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

$$x^{2}+5x+6=0$$

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.

$$a+b=5 ab=1\times 6=6$$

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 5.

$$a=2 b=3$$

Rewrite x^{2}+5x+6 as \left(x^{2}+2x\right)+\left(3x+6\right).

$$\left(x^{2}+2x\right)+\left(3x+6\right)$$

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

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

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

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

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

$$x=-2 x=-3$$

Steps Using the Quadratic Formula

Add 5x to both sides.

$$x^{2}+6+5x=0$$

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.

$$x^{2}+5x+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 5 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\times 6}}{2}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\times 6}}{2}$$

Multiply -4 times 6.

$$x=\frac{-5±\sqrt{25-24}}{2}$$

Add 25 to -24.

$$x=\frac{-5±\sqrt{1}}{2}$$

Take the square root of 1.

$$x=\frac{-5±1}{2}$$

Now solve the equation x=\frac{-5±1}{2} when ± is plus. Add -5 to 1.

$$x=-\frac{4}{2}$$

Divide -4 by 2.

$$x=-2$$

Now solve the equation x=\frac{-5±1}{2} when ± is minus. Subtract 1 from -5.

$$x=-\frac{6}{2}$$

Divide -6 by 2.

$$x=-3$$

The equation is now solved.

$$x=-2 x=-3$$

Steps for Completing the Square

Add 5x to both sides.

$$x^{2}+6+5x=0$$

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

$$x^{2}+5x=-6$$

Divide 5, the coefficient of the x term, by 2 to get \frac{5}{2}. Then add the square of \frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+5x+\left(\frac{5}{2}\right)^{2}=-6+\left(\frac{5}{2}\right)^{2}$$

Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+5x+\frac{25}{4}=-6+\frac{25}{4}$$

Add -6 to \frac{25}{4}.

$$x^{2}+5x+\frac{25}{4}=\frac{1}{4}$$

Factor x^{2}+5x+\frac{25}{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}.

$$\left(x+\frac{5}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$x+\frac{5}{2}=\frac{1}{2} x+\frac{5}{2}=-\frac{1}{2}$$

Subtract \frac{5}{2} from both sides of the equation.

$$x=-2 x=-3$$