How do you find the roots, real and imaginary, of #y=-15^2 + 12x +34 # using the quadratic formula?

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Answer 1 · 2017-07-19T14:12:00

See a solution process below:

(Assuming the problem is: #y = -15x^2 + 12x + 34#)

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

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

Substituting #-15# for #a#; #12# for #b# and #34# for #c# gives:

#x = (-12 +- sqrt(12^2 - (4 * -15 * 34)))/(2 * -15)#

#x = (-12 +- sqrt(144 - (-2040)))/(-30)#

#x = (-12 +- sqrt(144 + 2040))/(-30)#

#x = (-12 +- sqrt(2184))/(-30)#

#x = (-12)/(-30) +- (sqrt(2184))/(-30)#

#x = 2/5 +- (sqrt(2184))/(-30)#

#x = 2/5 +- (sqrt(4 * 546))/(-30)#

#x = 2/5 +- (sqrt(4)sqrt(546))/(-30)#

#x = 2/5 +- (2sqrt(546))/(-30)#

#x = 2/5 +- (sqrt(546))/(-15)#

#x = 2/5 - (sqrt(546))/(15)# and #x = 2/5 + (sqrt(546))/(15)#