How do you find the roots, real and imaginary, of #y=-3x^2 + 12x +14 # using the quadratic formula?

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Answer 1 · 2018-08-01T09:52:19

#x=-0.94# and #x=4.94#

The general form of a quadratic equation is #y=ax^2+bx+c#

The quadratic formula is #x=(-b+-sqrt(b^2-4ac))/(2a)# where #a,b,c# are from the general form

Looking at #y=-3x^2+12x+14#,
#a=-3#
#b=12#
#c=14#

#x=(-b+-sqrt(b^2-4ac))/(2a)#

#x=(-12+-sqrt(144+168))/(2times-3)#

#x=(-12+-sqrt312)/-6#

#x=(-12+sqrt312)/-6# or #x=(-12-sqrt312)/-6#

Therefore, the roots are #x=-0.94# and #x=4.94#
(both answers to 2.d.p)