How do you find the roots, real and imaginary, of #y=2x^2 -12x -5# using the quadratic formula?

1 Answer
Bill K.
Dec 7, 2015

There are two real roots that are irrational: #x=3 pm 1/2 sqrt(46)#

Explanation:

The quadratic formula says the roots of the function #y=ax^2+bx+c# are #x=(-b pm sqrt(b^{2}-4ac))/(2a)#.

For the present problem, #a=2#, #b=-12#, and #c=-5#. Hence, the roots are

#(12 pm sqrt{144-4*2*(-5)})/(2*2)=3 pm 1/4 sqrt{184}#

#=3 pm 1/4 sqrt{4}sqrt{46}=3 pm 1/2 sqrt(46)#.

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