100=15t^2+3

Solution for 100=15t^2+3 equation:

100=15t^2+3

We move all terms to the left:

100-(15t^2+3)=0

We get rid of parentheses

-15t^2-3+100=0

We add all the numbers together, and all the variables

-15t^2+97=0

a = -15; b = 0; c = +97;
Δ = b2-4ac
Δ = 02-4·(-15)·97
Δ = 5820
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:

$\sqrt{\Delta}=\sqrt{5820}=\sqrt{4*1455}=\sqrt{4}*\sqrt{1455}=2\sqrt{1455}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{1455}}{2*-15}=\frac{0-2\sqrt{1455}}{-30}
=-\frac{2\sqrt{1455}}{-30}
=-\frac{\sqrt{1455}}{-15}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{1455}}{2*-15}=\frac{0+2\sqrt{1455}}{-30}
=\frac{2\sqrt{1455}}{-30}
=\frac{\sqrt{1455}}{-15}
$