16t2-100t+15=0

Solution for 16t2-100t+15=0 equation:

16t^2-100t+15=0

a = 16; b = -100; c = +15;
Δ = b2-4ac
Δ = -1002-4·16·15
Δ = 9040
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:

$\sqrt{\Delta}=\sqrt{9040}=\sqrt{16*565}=\sqrt{16}*\sqrt{565}=4\sqrt{565}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-100)-4\sqrt{565}}{2*16}=\frac{100-4\sqrt{565}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-100)+4\sqrt{565}}{2*16}=\frac{100+4\sqrt{565}}{32}
$