(2h^2-5h)+(7h-3h^2)=0

Solution for (2h^2-5h)+(7h-3h^2)=0 equation:

(2h^2-5h)+(7h-3h^2)=0

We get rid of parentheses

-3h^2+2h^2+7h-5h=0

We add all the numbers together, and all the variables

-1h^2+2h=0

a = -1; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·(-1)·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*-1}=\frac{-4}{-2}
=+2
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*-1}=\frac{0}{-2}
=0
$