$a)\frac{1}{2}\cdot \left(x+5\right)\cdot \left(2x-4\right)=x^2$

$\left(x+5\right)\cdot \left(\frac{1}{2}\cdot \left(2x-4\right)\right)=x^2$

$\left(x+5\right)\cdot \left(x-2\right)=x^2$

$x\left(x-2\right)+5\left(x-2\right)=x^2$

$x^2-2x+5x-10=x^2$

$x^2+3x-10=x^2\mid -x^2$

$3x-10=0\mid +10$

$x=\frac{10}{3}=3\frac{1}{3}$

 

 

$b)\left(x-2\right)\cdot \left(x+10\right)=x^2$

$x\left(x+10\right)-2\left(x+10\right)=x^2$

$x^2+10x-2x-20=x^2$

$x^2+8x-20=x^2\mid -x^2$

$8x-20=0\mid +20$

$x=\frac{20}{8}=\frac{10}{4}=\frac{5}{2}=2\frac{1}{2}$

 

 

$c)\left(5+\left[\left(x-5\right)+\left(x-4\right)\right]\right)\cdot \left(x+3\right)\cdot \frac{1}{2}=x^2$

$\left(5+x-5+x-4\right)\cdot \left(x+3\right)\cdot \frac{1}{2}=x^2$

$\left(2x-4\right)\cdot \left(x+3\right)\cdot \frac{1}{2}=x^2$

$\left(\frac{1}{2}\cdot \left(2x-4\right)\right)\cdot \left(x+3\right)=x^2$

$\left(x-2\right)\cdot \left(x+3\right)=x^2$

$x\left(x+3\right)-2\left(x+3\right)=x^2$

$x^2+3x-2x-6=x^2$

$x^2+x-6=x^2\mid -x^2$

$x-6=0\mid +6$

$x=6$