# How do you solve 2t^2 – 6t + 1 = 0 using the quadratic formula?

Oct 14, 2015

The solutions are:
color(blue)(t=((3+sqrt(7)))/2

color(blue)(t=((3-sqrt(7)))/2

#### Explanation:

$2 {t}^{2} - 6 t + 1$

The equation is of the form color(blue)(at^2+bt+c=0 where:
$a = 2 , b = - 6 , c = 1$

The Discriminant is given by:

$\Delta = {b}^{2} - 4 \cdot a \cdot c$
$= {\left(- 6\right)}^{2} - \left(4 \cdot 2 \cdot 1\right)$
$= 36 - 8 = 28$

The solutions are found using the formula
$t = \frac{- b \pm \sqrt{\Delta}}{2 \cdot a}$

$t = \frac{- \left(- 6\right) \pm \sqrt{28}}{2 \cdot 2} = \frac{6 \pm 2 \sqrt{7}}{4}$

$\frac{6 \pm 2 \sqrt{7}}{4} = \frac{2 \left(3 \pm \sqrt{7}\right)}{4}$

$= \frac{\cancel{2} \left(3 \pm \sqrt{7}\right)}{\cancel{4}}$

$= \frac{\left(3 \pm \sqrt{7}\right)}{2}$

The solutions are
color(blue)(t=((3+sqrt(7)))/2

color(blue)(t=((3-sqrt(7)))/2