How do you find the roots, real and imaginary, of #y= -3x^2-16x +8 # using the quadratic formula?

1 Answer
Oct 22, 2017

Answer:

Roots are #**5.7936, -0.4603**#

Explanation:

Standard form of equation is #ax^2 + bx + c = 0#
Roots are# = ((-b) +- sqrt(b^2 -(4ac)))/(2a)#

Given equation is #-3x^2 -16x + 8 = 0#
#a = -3, b = -16, c = 8#

# x =( -(-16) +- sqrt((-16)^2 - (4*(-3)8)))/(2*(-3)) #

#x = (16 +- sqrt(256 + 96))/6#

#x = (16 +- sqrt352)/6#

#x = 5.7936, -0.4603#