u^2+22u+40=0

Solution for u^2+22u+40=0 equation:

u^2+22u+40=0

a = 1; b = 22; c = +40;
Δ = b2-4ac
Δ = 222-4·1·40
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-18}{2*1}=\frac{-40}{2}
=-20
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+18}{2*1}=\frac{-4}{2}
=-2
$