# How do you solve x^2-5x=14?

Mar 22, 2017

Set up two binomials.
The sum of the two numbers must be -5 and the product -14.
Then solve for x

#### Explanation:

${x}^{2} - 5 x = 14$

Subtract 14 from both sides of the equation to equal 0

${x}^{2} - 5 x - 14 = 14 - 14 \text{ }$ (14-14 = 0) giving

${x}^{2} - 5 x - 14 = 0$
The -14 indicates that one number must be positive and one number must be negative.

The -5 indicates that the negative number must be larger than the positive number.

Factor -14 The possibilities are $- 2 \times + 7 , + 2 \times - 7 , - 1 \times + 14 , + 1 \times - 14$
The combination that works is $+ 2 \times - 7 ,$ so the binomials are

$\left(x + 2\right) \times \left(x - 7\right) = 0$

Now set both binomials equal to zero and solve for x

$x + 2 = 0 \text{ }$ Subtract 2 from both sides

$x + 2 - 2 = 0 - 2 \text{ } \left(2 - 2 = 0\right)$ so

$x = - 2$

$x - 7 = 0 \text{ }$ add 7 to both sides

$x - 7 + 7 = 0 + 7 \text{ } \left(- 7 + 7 = 0\right)$ so

$x = + 7$