16t^2+100t+85=0

Solution for 16t^2+100t+85=0 equation:

16t^2+100t+85=0

a = 16; b = 100; c = +85;
Δ = b2-4ac
Δ = 1002-4·16·85
Δ = 4560
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:

$\sqrt{\Delta}=\sqrt{4560}=\sqrt{16*285}=\sqrt{16}*\sqrt{285}=4\sqrt{285}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(100)-4\sqrt{285}}{2*16}=\frac{-100-4\sqrt{285}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(100)+4\sqrt{285}}{2*16}=\frac{-100+4\sqrt{285}}{32}
$