Solve x^2+10x=56

$$x ^ { 2 } + 10 x = 56$$

Solve for x

$$x=-14$$

$$x=4$$

Steps Using Factoring

Subtract 56 from both sides.

$$x^{2}+10x-56=0$$

To solve the equation, factor x^{2}+10x-56 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.

$$a+b=10 ab=-56$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -56.

$$-1,56 -2,28 -4,14 -7,8$$

Calculate the sum for each pair.

$$-1+56=55 -2+28=26 -4+14=10 -7+8=1$$

The solution is the pair that gives sum 10.

$$a=-4 b=14$$

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

$$\left(x-4\right)\left(x+14\right)$$

To find equation solutions, solve x-4=0 and x+14=0.

$$x=4 x=-14$$

Steps Using Factoring By Grouping

Subtract 56 from both sides.

$$x^{2}+10x-56=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-56. To find a and b, set up a system to be solved.

$$a+b=10 ab=1\left(-56\right)=-56$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -56.

$$-1,56 -2,28 -4,14 -7,8$$

Calculate the sum for each pair.

$$-1+56=55 -2+28=26 -4+14=10 -7+8=1$$

The solution is the pair that gives sum 10.

$$a=-4 b=14$$

Rewrite x^{2}+10x-56 as \left(x^{2}-4x\right)+\left(14x-56\right).

$$\left(x^{2}-4x\right)+\left(14x-56\right)$$

Factor out x in the first and 14 in the second group.

$$x\left(x-4\right)+14\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(x+14\right)$$

To find equation solutions, solve x-4=0 and x+14=0.

$$x=4 x=-14$$

Steps Using the Quadratic Formula

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.

$$x^{2}+10x=56$$

Subtract 56 from both sides of the equation.

$$x^{2}+10x-56=56-56$$

Subtracting 56 from itself leaves 0.

$$x^{2}+10x-56=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 10 for b, and -56 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\left(-56\right)}}{2}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\left(-56\right)}}{2}$$

Multiply -4 times -56.

$$x=\frac{-10±\sqrt{100+224}}{2}$$

Add 100 to 224.

$$x=\frac{-10±\sqrt{324}}{2}$$

Take the square root of 324.

$$x=\frac{-10±18}{2}$$

Now solve the equation x=\frac{-10±18}{2} when ± is plus. Add -10 to 18.

$$x=\frac{8}{2}$$

Divide 8 by 2.

$$x=4$$

Now solve the equation x=\frac{-10±18}{2} when ± is minus. Subtract 18 from -10.

$$x=-\frac{28}{2}$$

Divide -28 by 2.

$$x=-14$$

The equation is now solved.

$$x=4 x=-14$$

Steps for Completing the Square

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$x^{2}+10x=56$$

Divide 10, the coefficient of the x term, by 2 to get 5. Then add the square of 5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+10x+5^{2}=56+5^{2}$$

Square 5.

$$x^{2}+10x+25=56+25$$

Add 56 to 25.

$$x^{2}+10x+25=81$$

Factor x^{2}+10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+5\right)^{2}=81$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+5\right)^{2}}=\sqrt{81}$$

Simplify.

$$x+5=9 x+5=-9$$

Subtract 5 from both sides of the equation.

$$x=4 x=-14$$