How do you solve q^2-21q=-20?

Apr 2, 2016

The solutions are:

 color(green)(q = 20

 color(green)(q = 1

Explanation:

${q}^{2} - 21 q = - 20$

${q}^{2} - 21 q + 20 = 0$

The equation is of the form color(blue)(aq^2+bq+c=0 where:

$a = 1 , b = - 21 , c = 20$

The Discriminant is given by:

$\Delta = {b}^{2} - 4 \cdot a \cdot c$

$= {\left(- 21\right)}^{2} - \left(4 \cdot 1 \cdot 20\right)$

$= 441 - 80 = 361$

The solutions are normally found using the formula:

$q = \frac{- b \pm \sqrt{\Delta}}{2 \cdot a}$

$q = \frac{- \left(- 21\right) \pm \sqrt{361}}{2 \cdot 1} = \frac{\left(21 \pm 19\right)}{2}$

$q = \frac{\left(21 + 19\right)}{2} = \frac{40}{2} = 20$

$q = \frac{\left(21 - 19\right)}{2} = \frac{2}{2} = 1$