Question

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

Answer 1

9-12х+4x^2-(4x^2+10x-10x-25)=0
9-12х+4x^2-(4x^2-25)=0
9-12х+4x^2-4x^2+25=0
9+25=12x
34=12x
x=34/12=17/6=2 (5/6)

Answer 2

$(3-2x)^2-(2x-5)(2x+5)=0\3^2-12x+(2x)^2-((2x)^2-5^2)=0\9-12x+4x^2-4x^2+25=0\-12x=-9-25\x=frac{-34}{-12}=frac{17}{6}=2frac{5}{6}$