Solve x^2-5x-10=-2x

$$x ^ { 2 } - 5 x - 10 = - 2 x$$

Solve for x

$$x=-2$$

$$x=5$$

Steps Using Factoring

Add 2x to both sides.

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

Combine -5x and 2x to get -3x.

$$x^{2}-3x-10=0$$

To solve the equation, factor x^{2}-3x-10 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.

$$a+b=-3 ab=-10$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -10.

$$1,-10 2,-5$$

Calculate the sum for each pair.

$$1-10=-9 2-5=-3$$

The solution is the pair that gives sum -3.

$$a=-5 b=2$$

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

$$\left(x-5\right)\left(x+2\right)$$

To find equation solutions, solve x-5=0 and x+2=0.

$$x=5 x=-2$$

Steps Using Factoring By Grouping

Add 2x to both sides.

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

Combine -5x and 2x to get -3x.

$$x^{2}-3x-10=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-10. To find a and b, set up a system to be solved.

$$a+b=-3 ab=1\left(-10\right)=-10$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -10.

$$1,-10 2,-5$$

Calculate the sum for each pair.

$$1-10=-9 2-5=-3$$

The solution is the pair that gives sum -3.

$$a=-5 b=2$$

Rewrite x^{2}-3x-10 as \left(x^{2}-5x\right)+\left(2x-10\right).

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

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

$$x\left(x-5\right)+2\left(x-5\right)$$

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

$$\left(x-5\right)\left(x+2\right)$$

To find equation solutions, solve x-5=0 and x+2=0.

$$x=5 x=-2$$

Steps Using the Quadratic Formula

Add 2x to both sides.

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

Combine -5x and 2x to get -3x.

$$x^{2}-3x-10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -3 for b, and -10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-10\right)}}{2}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-10\right)}}{2}$$

Multiply -4 times -10.

$$x=\frac{-\left(-3\right)±\sqrt{9+40}}{2}$$

Add 9 to 40.

$$x=\frac{-\left(-3\right)±\sqrt{49}}{2}$$

Take the square root of 49.

$$x=\frac{-\left(-3\right)±7}{2}$$

The opposite of -3 is 3.

$$x=\frac{3±7}{2}$$

Now solve the equation x=\frac{3±7}{2} when ± is plus. Add 3 to 7.

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

Divide 10 by 2.

$$x=5$$

Now solve the equation x=\frac{3±7}{2} when ± is minus. Subtract 7 from 3.

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

Divide -4 by 2.

$$x=-2$$

The equation is now solved.

$$x=5 x=-2$$

Steps for Completing the Square

Add 2x to both sides.

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

Combine -5x and 2x to get -3x.

$$x^{2}-3x-10=0$$

Add 10 to both sides. Anything plus zero gives itself.

$$x^{2}-3x=10$$

Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=10+\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=10+\frac{9}{4}$$

Add 10 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{49}{4}$$

Factor x^{2}-3x+\frac{9}{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}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

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

Add \frac{3}{2} to both sides of the equation.

$$x=5 x=-2$$