16t^2-36t+15=0

Solution for 16t^2-36t+15=0 equation:

16t^2-36t+15=0

a = 16; b = -36; c = +15;
Δ = b2-4ac
Δ = -362-4·16·15
Δ = 336
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:

$\sqrt{\Delta}=\sqrt{336}=\sqrt{16*21}=\sqrt{16}*\sqrt{21}=4\sqrt{21}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-4\sqrt{21}}{2*16}=\frac{36-4\sqrt{21}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+4\sqrt{21}}{2*16}=\frac{36+4\sqrt{21}}{32}
$