56t^2+t-57=0

Solution for 56t^2+t-57=0 equation:

56t^2+t-57=0

a = 56; b = 1; c = -57;
Δ = b2-4ac
Δ = 12-4·56·(-57)
Δ = 12769
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{12769}=113$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-113}{2*56}=\frac{-114}{112}
=-1+1/56
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+113}{2*56}=\frac{112}{112}
=1
$