-2u^2-3u+14=0

Solution for -2u^2-3u+14=0 equation:

-2u^2-3u+14=0

a = -2; b = -3; c = +14;
Δ = b2-4ac
Δ = -32-4·(-2)·14
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{-(-3)-11}{2*-2}=\frac{-8}{-4}
=+2
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+11}{2*-2}=\frac{14}{-4}
=-3+1/2
$