# How do you solve x^ { 2} - 11x + 19= - 5?

Dec 5, 2017

$x = 3 , 8$

#### Explanation:

STEP ONE: Add $5$ to each side. You want all of the terms to be on one side before you start factoring!

${x}^{2} - 11 x + 19 = - 5$

${x}^{2} - 11 x + 24 = 0$

STEP TWO: Factor the equation! Two number that equal $24$ when multiplied and $- 11$ when added together are $- 8$ and $- 3$

${x}^{2} - 11 x + 24 = 0$

$\left(x - 3\right) \left(x - 8\right)$

THEREFORE, $x = 3 , 8$

Dec 5, 2017

$3 = x = 8$

#### Explanation:

First, we turn the equation so that it equals zero.
${x}^{2} - 11 x + 19 = - 5$
${x}^{2} - 11 x + 24 = 0$

Let's see whether we could factor this.

We find the factors for 24:
1,24
2,12
3,8
4,6
-1,-24
-2,-12
-3,-8
-4,-6

We see that -3 and -8 add up to -11.

Therefore, ${x}^{2} - 11 x + 24 = 0$ becomes $\left(x - 3\right) \left(x - 8\right) = 0$
Therefore, our answers are 3 and 8.

You could have used the quadratic formula, but it is unnecessary in this case.

Dec 5, 2017

$2$ real solutions:

$x = 3 \setminus \quad , \setminus \quad x = 8$

#### Explanation:

Rearrange the equation so it becomes a standard form quadratic equation, $a {x}^{2} + b x + c$.

${x}^{2} - 11 x + 24$, where:

• $a = 1$
• $b = - 11$
• $c = 24$

Plug those values into the quadratic formula:

$x = \setminus \frac{- b \setminus \pm \setminus \sqrt{{b}^{2} - 4 a c}}{2 a}$

$\setminus \implies x = \setminus \frac{11 \setminus \pm \setminus \sqrt{{\left(- 11\right)}^{2} - 4 \left(1\right) \left(24\right)}}{2 \left(1\right)}$

$\setminus \implies x = \setminus \frac{11 \setminus \pm \setminus \sqrt{121 - 96}}{2}$

$\setminus \implies x - \setminus \frac{11 \setminus \pm 5}{2}$

$\setminus \implies x = 3 \setminus \quad , \setminus \quad x = 8$