How do you solve #2x^2+3x-5=0# using the quadratic formula?

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Answer 1 · 2015-08-10T07:19:25

The solutions for the equation are:
#color(blue)(x=-5/2, x=1#

#2x^2+3x−5=0#

The equation is of the form #color(blue)(ax^2+bx+c=0# where:
#a=2, b=3, c=-5#

The Discriminant is given by:
#Delta=b^2-4*a*c#

# = (3)^2-(4*2* (-5))#

# = 9+40#

#=49#

The solutions are found using the formula
#x=(-b+-sqrtDelta)/(2*a)#

#x = ((-3)+-sqrt(49))/(2*2) = ((-3+-7))/4#

#x = (-3-7)/4, color(blue)(x=-5/2#

#x=(-3+7)/4, color(blue)(x=1#