16t^2+15t-25=0

Solution for 16t^2+15t-25=0 equation:

16t^2+15t-25=0

a = 16; b = 15; c = -25;
Δ = b2-4ac
Δ = 152-4·16·(-25)
Δ = 1825
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:

$\sqrt{\Delta}=\sqrt{1825}=\sqrt{25*73}=\sqrt{25}*\sqrt{73}=5\sqrt{73}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-5\sqrt{73}}{2*16}=\frac{-15-5\sqrt{73}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+5\sqrt{73}}{2*16}=\frac{-15+5\sqrt{73}}{32}
$