(u^2-6u+2)+(-3u^2+7u+5)=0

Solution for (u^2-6u+2)+(-3u^2+7u+5)=0 equation:

(u^2-6u+2)+(-3u^2+7u+5)=0

We get rid of parentheses

-3u^2+u^2+7u-6u+5+2=0

We add all the numbers together, and all the variables

-2u^2+u+7=0

a = -2; b = 1; c = +7;
Δ = b2-4ac
Δ = 12-4·(-2)·7
Δ = 57
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{57}}{2*-2}=\frac{-1-\sqrt{57}}{-4}
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{57}}{2*-2}=\frac{-1+\sqrt{57}}{-4}
$