# How do you solve t^2 + 13t + 42 = 0 using the quadratic formula?

Apr 26, 2016

The solutions for the equation are:
color(green)( t =-6, color(green)( t =-7

#### Explanation:

${t}^{2} + 13 t + 42 = 0$

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

$a = 1 , b = 13 , c = 42$

The Discriminant is given by:

color(blue)(Delta=b^2-4*a*c

$= {\left(13\right)}^{2} - \left(4 \cdot 1 \cdot 42\right)$

$= 169 - 168 = 1$

The solutions are found using the formula
color(blue)(t=(-b+-sqrtDelta)/(2*a)

$t = \frac{\left(- 13\right) \pm \sqrt{1}}{2 \cdot 1} = \frac{- 13 \pm 1}{2}$

$t = \frac{- 13 + 1}{2} = - \frac{12}{2} = - 6$

$t = \frac{- 13 - 1}{2} = - \frac{14}{2} = - 7$

The solutions are:

• color(green)( t =-6
• color(green)( t =-7