15t^2-8=0

Solution for 15t^2-8=0 equation:

15t^2-8=0

a = 15; b = 0; c = -8;
Δ = b2-4ac
Δ = 02-4·15·(-8)
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*15}=\frac{0-4\sqrt{30}}{30}
=-\frac{4\sqrt{30}}{30}
=-\frac{2\sqrt{30}}{15}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{30}}{2*15}=\frac{0+4\sqrt{30}}{30}
=\frac{4\sqrt{30}}{30}
=\frac{2\sqrt{30}}{15}
$