56t^2-19t-15=0

Solution for 56t^2-19t-15=0 equation:

56t^2-19t-15=0

a = 56; b = -19; c = -15;
Δ = b2-4ac
Δ = -192-4·56·(-15)
Δ = 3721
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3721}=61$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-61}{2*56}=\frac{-42}{112}
=-3/8
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+61}{2*56}=\frac{80}{112}
=5/7
$