How do you solve #5x^2+21x=-18# using the quadratic formula?

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Answer 1 · 2016-05-24T15:21:48

The solutions for the equation are:

#color(blue)(x =-6/5#

#color(blue)(x=-3#

#5x^2 + 21x = - 18#

#5x^2 + 21x + 18 = 0#

The equation is of the form #color(blue)(ax^2+bx+c=0# where:
#a=5, b=21, c=18#

The Discriminant is given by:

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

# = (21)^2-(4* 5 * 18)#

# = 441-360 = 81#

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

#x = ((-21)+-sqrt(81))/(2*5) = (-21+-9)/10#

#x = (-21 + 9 ) / 10 = -12 / 10 , color(blue)(x =-6/5#

#x = (-21 - 9 ) / 10 = -30 / 10 , color(blue)(x=-3#