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

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Answer 1 · 2016-03-06T14:56:40

The solutions are:
#x = color(green)(3/5#

#x =color(green)(- 1#

#5x^2 +2x -3=0#

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

The Discriminant is given by:

#Delta=b^2-4*a*c#

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

# = 4 +60 = 64#

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

#x = ((-2)+-sqrt(64))/(2*5) = (-2+-8)/10#

#x= (-2+8)/10= 6/10 = color(green)(3/5#

#x= (-2-8)/10= -10/10 =color(green)(- 1#