15=2.38t^2

Solution for 15=2.38t^2 equation:

15=2.38t^2

We move all terms to the left:

15-(2.38t^2)=0

We get rid of parentheses

-2.38t^2+15=0

a = -2.38; b = 0; c = +15;
Δ = b2-4ac
Δ = 02-4·(-2.38)·15
Δ = 142.8
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-\sqrt{142.8}}{2*-2.38}=\frac{0-\sqrt{142.8}}{-4.76}
=-\frac{\sqrt{}}{-4.76}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{142.8}}{2*-2.38}=\frac{0+\sqrt{142.8}}{-4.76}
=\frac{\sqrt{}}{-4.76}
$