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: