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

1 Answer
Aug 1, 2018

x=-0.94 and x=4.94

Explanation:

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)