100=(5t)^2+5

Solution for 100=(5t)^2+5 equation:

100=(5t)^2+5

We move all terms to the left:

100-((5t)^2+5)=0

We get rid of parentheses

-5t^2-5+100=0

We add all the numbers together, and all the variables

-5t^2+95=0

a = -5; b = 0; c = +95;
Δ = b2-4ac
Δ = 02-4·(-5)·95
Δ = 1900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1900}=\sqrt{100*19}=\sqrt{100}*\sqrt{19}=10\sqrt{19}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{19}}{2*-5}=\frac{0-10\sqrt{19}}{-10}
=-\frac{10\sqrt{19}}{-10}
=-\frac{\sqrt{19}}{-1}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{19}}{2*-5}=\frac{0+10\sqrt{19}}{-10}
=\frac{10\sqrt{19}}{-10}
=\frac{\sqrt{19}}{-1}
$