u^2-40u+400=16

Solution for u^2-40u+400=16 equation:

u^2-40u+400=16

We move all terms to the left:

u^2-40u+400-(16)=0

We add all the numbers together, and all the variables

u^2-40u+384=0

a = 1; b = -40; c = +384;
Δ = b2-4ac
Δ = -402-4·1·384
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-8}{2*1}=\frac{32}{2}
=16
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+8}{2*1}=\frac{48}{2}
=24
$