# How do you find the solution to the quadratic equation 2(x^2 - 5)^2 - 13 (x^2 - 5) + 20 = 0?

May 1, 2015

Let $p = \left({x}^{2} - 5\right)$
and solve
$2 {p}^{2} - 13 p + 20 = 0$
Then for each solution (in terms of $p$)
solve for $x$

$2 {p}^{2} - 13 p + 20 = 0$
$\rightarrow \left(p - 4\right) \left(2 p - 5\right) = 0$
$\rightarrow p = 4 \text{ or } p = \frac{5}{2}$

If $p = 4$
${x}^{2} - 5 = 4$
$\rightarrow x = + 3 \text{ or } x = - 3$

If $p = \frac{5}{2}$
${x}^{2} - 5 = \frac{5}{2}$
$\rightarrow x = + \frac{15}{2} = + 7 \frac{1}{2} \text{ or } x = - \frac{15}{2} = - 7 \frac{1}{2}$

So
$x \epsilon \left\{- 7 \frac{1}{2} , - 3 , + 3 , + 7 \frac{1}{2}\right\}$