15=8t^2

Solution for 15=8t^2 equation:

15=8t^2

We move all terms to the left:

15-(8t^2)=0

a = -8; b = 0; c = +15;
Δ = b2-4ac
Δ = 02-4·(-8)·15
Δ = 480
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:

$\sqrt{\Delta}=\sqrt{480}=\sqrt{16*30}=\sqrt{16}*\sqrt{30}=4\sqrt{30}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{30}}{2*-8}=\frac{0-4\sqrt{30}}{-16}
=-\frac{4\sqrt{30}}{-16}
=-\frac{\sqrt{30}}{-4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{30}}{2*-8}=\frac{0+4\sqrt{30}}{-16}
=\frac{4\sqrt{30}}{-16}
=\frac{\sqrt{30}}{-4}
$