(2x+3x−3)−25(x−32x+3)=5
Let2x+3x−3=y
2y−25×1y=5
2y2−25=5y
2y2−25−25=0
2y2+5y−1y−25=0
y(2y+5)−5(2y+5)=0
(2y+5)(y−5)=0
y=−52,5
ify=−52
2x+3x−3=−52
4x+6=−5x+5
9x=9
x=1
ify=5
2x+3x−3=5
2x+3=5x−15$
3x=18
x=6
Therefore the roots are real and distinct