Solve x^2+10x+5x+50=0

$$x ^ { 2 } + 10 x + 5 x + 50 = 0$$

Solve for x

$$x=-10$$

$$x=-5$$

Steps Using Factoring

Combine 10x and 5x to get 15x.

$$x^{2}+15x+50=0$$

To solve the equation, factor x^{2}+15x+50 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.

$$a+b=15 ab=50$$

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 50.

$$1,50 2,25 5,10$$

Calculate the sum for each pair.

$$1+50=51 2+25=27 5+10=15$$

The solution is the pair that gives sum 15.

$$a=5 b=10$$

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

$$\left(x+5\right)\left(x+10\right)$$

To find equation solutions, solve x+5=0 and x+10=0.

$$x=-5 x=-10$$

Steps Using Factoring By Grouping

Combine 10x and 5x to get 15x.

$$x^{2}+15x+50=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+50. To find a and b, set up a system to be solved.

$$a+b=15 ab=1\times 50=50$$

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 50.

$$1,50 2,25 5,10$$

Calculate the sum for each pair.

$$1+50=51 2+25=27 5+10=15$$

The solution is the pair that gives sum 15.

$$a=5 b=10$$

Rewrite x^{2}+15x+50 as \left(x^{2}+5x\right)+\left(10x+50\right).

$$\left(x^{2}+5x\right)+\left(10x+50\right)$$

Factor out x in the first and 10 in the second group.

$$x\left(x+5\right)+10\left(x+5\right)$$

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

$$\left(x+5\right)\left(x+10\right)$$

To find equation solutions, solve x+5=0 and x+10=0.

$$x=-5 x=-10$$

Steps Using the Quadratic Formula

Combine 10x and 5x to get 15x.

$$x^{2}+15x+50=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 15 for b, and 50 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-15±\sqrt{15^{2}-4\times 50}}{2}$$

Square 15.

$$x=\frac{-15±\sqrt{225-4\times 50}}{2}$$

Multiply -4 times 50.

$$x=\frac{-15±\sqrt{225-200}}{2}$$

Add 225 to -200.

$$x=\frac{-15±\sqrt{25}}{2}$$

Take the square root of 25.

$$x=\frac{-15±5}{2}$$

Now solve the equation x=\frac{-15±5}{2} when ± is plus. Add -15 to 5.

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

Divide -10 by 2.

$$x=-5$$

Now solve the equation x=\frac{-15±5}{2} when ± is minus. Subtract 5 from -15.

$$x=-\frac{20}{2}$$

Divide -20 by 2.

$$x=-10$$

The equation is now solved.

$$x=-5 x=-10$$

Steps for Completing the Square

Combine 10x and 5x to get 15x.

$$x^{2}+15x+50=0$$

Subtract 50 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}+15x=-50$$

Divide 15, the coefficient of the x term, by 2 to get \frac{15}{2}. Then add the square of \frac{15}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+15x+\left(\frac{15}{2}\right)^{2}=-50+\left(\frac{15}{2}\right)^{2}$$

Square \frac{15}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+15x+\frac{225}{4}=-50+\frac{225}{4}$$

Add -50 to \frac{225}{4}.

$$x^{2}+15x+\frac{225}{4}=\frac{25}{4}$$

Factor x^{2}+15x+\frac{225}{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}.

$$\left(x+\frac{15}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{15}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$x+\frac{15}{2}=\frac{5}{2} x+\frac{15}{2}=-\frac{5}{2}$$

Subtract \frac{15}{2} from both sides of the equation.

$$x=-5 x=-10$$