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

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Answer 1 · 2016-04-26T14:35:29

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

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

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:

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

$= (13)^2-(4* 1 * 42)$

$= 169 - 168 = 1$

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

$t = ((-13)+-sqrt(1))/(2*1) = (-13+- 1)/2$

$t = (-13+ 1)/2 = -12 /2 = -6$

$t = (-13- 1)/2 = -14/2 = -7$

The solutions are:

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