48t^2-14t-49=0

Solution for 48t^2-14t-49=0 equation:

48t^2-14t-49=0

a = 48; b = -14; c = -49;
Δ = b2-4ac
Δ = -142-4·48·(-49)
Δ = 9604
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9604}=98$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-98}{2*48}=\frac{-84}{96}
=-7/8
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+98}{2*48}=\frac{112}{96}
=1+1/6
$