16t^2-14t-15=0

Solution for 16t^2-14t-15=0 equation:

16t^2-14t-15=0

a = 16; b = -14; c = -15;
Δ = b2-4ac
Δ = -142-4·16·(-15)
Δ = 1156
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1156}=34$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-34}{2*16}=\frac{-20}{32}
=-5/8
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+34}{2*16}=\frac{48}{32}
=1+1/2
$