How do you find the roots, real and imaginary, of #y=(4x-5)(x+1)# using the quadratic formula?

Question

1711 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-16T19:38:35

#x_1=5/4#
#x_2=-1#

#(4x-5)*(x+1)#

#=4x^2+4x*1-5*x-5#

#=4x^2+4x-5x-5#

#=4x^2-x-5=#

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

#x_(1,2)=(1+-sqrt(1+80))/8#

#x_(1,2)=(1+-9)/8#

#x_1=5/4#

#x_2=-1#