How do you find the roots, real and imaginary, of #y=-5x^2 +x -4 # using the quadratic formula?

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Answer 1 · 2017-07-16T00:05:10

See a solution process below:

For #ax^2 + bx + c = 0#, the values of #x# which are the solutions to the equation are given by:

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

Substituting #-5# for #a#; #1# for #b# and #-4# for #c# gives:

#x = (-1 +- sqrt(1^2 - (4 * -5 * -4)))/(2 * -5)#

#x = (-1 +- sqrt(1 - (80)))/(-10)#

#x = (-1 +- sqrt(-79))/(-10)#

#x = (-1 + sqrt(-79))/(-10)# and #x = (-1 - sqrt(-79))/(-10)#