(5u^2)+(-50)=01

Solution for (5u^2)+(-50)=01 equation:

(5u^2)+(-50)=01

We move all terms to the left:

(5u^2)+(-50)-(01)=0

determiningTheFunctionDomain
5u^2-01+(-50)=0

We add all the numbers together, and all the variables

5u^2-51=0

a = 5; b = 0; c = -51;
Δ = b2-4ac
Δ = 02-4·5·(-51)
Δ = 1020
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:

$\sqrt{\Delta}=\sqrt{1020}=\sqrt{4*255}=\sqrt{4}*\sqrt{255}=2\sqrt{255}$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{255}}{2*5}=\frac{0-2\sqrt{255}}{10}
=-\frac{2\sqrt{255}}{10}
=-\frac{\sqrt{255}}{5}
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{255}}{2*5}=\frac{0+2\sqrt{255}}{10}
=\frac{2\sqrt{255}}{10}
=\frac{\sqrt{255}}{5}
$