# Question #6158f

May 27, 2017

$n = 6 \mathmr{and} n = 4$

#### Explanation:

• The first step with a quadratic equation is to make it equal to $0$

${n}^{2} - 10 n + 22 = - 2$

${n}^{2} - 10 n + 24 = 0$

• The second step is to factorise if possible.

The $+$ with $24$ tells you that you must find factors of $24$ which ADD up to $10$

$6 \mathmr{and} 4$ will work: $6 \times 4 = 24 \text{ and } 6 + 4 = 10$

The $+$ with $24$ also tells you that the signs of the factors must be the same.

The $-$ with $10$ tells you that the signs will both be $\text{negative}$

${n}^{2} - 10 n + 24 = 0$

$\left(n - 6\right) \left(n - 4\right) = 0$

• The third step is to set each factor equal to $0$ and solve it.

$n - 6 = 0 \text{ } \rightarrow n = 6$
$n - 4 = 0 \text{ } \rightarrow n = 4$

Jul 8, 2017

4, and 6

#### Explanation:

$f \left(n\right) = {n}^{2} - 10 n + 24 = 0$
Use the new Transforming Method (Google Search), when a = 1.
Find 2 real roots, that have same sign (ac > 0), knowing the sum (-b = 10) and the product (c = 24). They are 4 and 6.