How do you find the roots, real and imaginary, of #y=12x^2 -7x -12 # using the quadratic formula?

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Answer 1 · 2018-05-10T01:57:48

Your roots are #x = -3/4# or #1 1/3#

Using the quadratic formula:
#x = (-b +- sqrt(b^2 - 4ac))/(2a)#

By inputting values, we're given:

#x = (-(-7) +- sqrt(7^2 - (4*12*-12)))/(2*12)#

#x = (7 +- sqrt(49 + 576))/24#

#x = (7 +- sqrt(625))/24#

#x = (7 +- 25)/24#

#x = (7 + 25)/24 = 32/24 = 1 1/3#

#x = (7 - 25)/24 = -18/24 = -3/4#

Your roots are #x = -3/4# or #1 1/3#