U3=14u2+32u

Solution for U3=14u2+32u equation:

3=14U^2+32U

We move all terms to the left:

3-(14U^2+32U)=0

We get rid of parentheses

-14U^2-32U+3=0

a = -14; b = -32; c = +3;
Δ = b2-4ac
Δ = -322-4·(-14)·3
Δ = 1192
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$U_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$U_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1192}=\sqrt{4*298}=\sqrt{4}*\sqrt{298}=2\sqrt{298}$
$U_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-2\sqrt{298}}{2*-14}=\frac{32-2\sqrt{298}}{-28}
$
$U_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+2\sqrt{298}}{2*-14}=\frac{32+2\sqrt{298}}{-28}
$