(U^2-7u-2)+(5u^2-4u+3)=0

Solution for (U^2-7u-2)+(5u^2-4u+3)=0 equation:

(^2-7U-2)+(5U^2-4U+3)=0

We get rid of parentheses

5U^2-7U-4U-2+3+^2=0

We add all the numbers together, and all the variables

5U^2-11U=0

a = 5; b = -11; c = 0;
Δ = b2-4ac
Δ = -112-4·5·0
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$U_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$U_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$U_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-11}{2*5}=\frac{0}{10}
=0
$
$U_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+11}{2*5}=\frac{22}{10}
=2+1/5
$