#4x^2+12x-25=ax^2+bx+c#
set #4x^2+12x-25# to 0 to find the roots real and imaginary
Using the discriminant #b^2-4ac# to find out if there are imaginary roots
#(12)^2-4(4)(-25)=544#
#Delta("the discriminant")>0# So that means it have 2 unique real roots
Using the quadratic formula
#x=(-(12)+-sqrt((12)^2-4(4)(-25)))/(2(4))#
#x=(-12+-sqrt544)/8#
simplify the radical using #root(n)xy=root(n)x*root(n)y#
#x=(-12+-sqrt(16*34))/8#
#x=(-12+-sqrt(4)sqrt(34))/8#
#x=(-12+-2sqrt(34))/8#
Find the greatest common factor
#x=(-3+-sqrt(34))/2#
#x=(-3+sqrt(34))/2# and #-(3+sqrt(34))/2#
Using #C.T.S.# (complete the square)
To get #x^2# by itself divide the whole equation by #4#
#(4x^2+12x-25)/4=0/4#
#(cancel4x^2)/cancel4+(3cancel12x)/(1cancel4)-25/4=0#
#x^2+3xcancel(-25/4+25/4)=0+25/4#
#x^2+3x=25/4#
complete the square by dividing the coefficient of x by 2 and then squaring it
#(3/2)^2=9/4#
add this term to both sides
#x^2+3x+9/4=25/4+9/4#
#sqrt((x+3/2)^2)=+-sqrt(34/4#
#xcancel(+3/2-3/2)=(+-sqrt(34))/2+(-3)/2#
#x=(-3+-sqrt34)/2#
#:.# #x=(-3+sqrt34)/2# and #(-3-sqrt34)/2#
