54=u^2+4

Solution for 54=u^2+4 equation:

54=u^2+4

We move all terms to the left:

54-(u^2+4)=0

We get rid of parentheses

-u^2-4+54=0

We add all the numbers together, and all the variables

-1u^2+50=0

a = -1; b = 0; c = +50;
Δ = b2-4ac
Δ = 02-4·(-1)·50
Δ = 200
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:

$\sqrt{\Delta}=\sqrt{200}=\sqrt{100*2}=\sqrt{100}*\sqrt{2}=10\sqrt{2}$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{2}}{2*-1}=\frac{0-10\sqrt{2}}{-2}
=-\frac{10\sqrt{2}}{-2}
=-\frac{5\sqrt{2}}{-1}
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{2}}{2*-1}=\frac{0+10\sqrt{2}}{-2}
=\frac{10\sqrt{2}}{-2}
=\frac{5\sqrt{2}}{-1}
$