2x^(2)+5x-12=0

3x^(2)-5x-12=0

If sinalphaandcosalpha are the roots of 25x^(2)+5x-12=0 then the value of sin2alpha is

A= {x//x^(2) -7x +12 =0}, B= {x//x^(2) -x-12=0}, then AnnB=

A= {x//x^(2) -7x +12 =0}, B= {x//x^(2) -x-12=0}, then AnnB=

If X=(x:x^(2)-12x+20=0) and Y=(x:x^(2)+5x-14=0) , then X-Y=

cosalpha is a root of the equation 25 x^2+5x-12=0 , -1 le x le 0 then the value of sin2alpha is:

Solve: ,x^(2)-1,x^(2)+2x+1,2x^(2)+3x+12x^(2)+x-1,2x^(2)+5x-3,2x^(2)+4x-36x^(2)-x-2,6x^(2)-7x+2,12x^(2)-5x-2]|=0

If sin alpha and cos alpha are the roots of 25 x ^2 +5x-12=0 , then value of sin 2alpha is

Integrate : int log (2x^(2)+5x-12)dx