-2u^2-12u+27=0

Solution for -2u^2-12u+27=0 equation:

-2u^2-12u+27=0

a = -2; b = -12; c = +27;
Δ = b2-4ac
Δ = -122-4·(-2)·27
Δ = 360
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:

$\sqrt{\Delta}=\sqrt{360}=\sqrt{36*10}=\sqrt{36}*\sqrt{10}=6\sqrt{10}$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-6\sqrt{10}}{2*-2}=\frac{12-6\sqrt{10}}{-4}
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+6\sqrt{10}}{2*-2}=\frac{12+6\sqrt{10}}{-4}
$