t^2-14t-48=0

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

t^2-14t-48=0

a = 1; b = -14; c = -48;
Δ = b2-4ac
Δ = -142-4·1·(-48)
Δ = 388
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:

$\sqrt{\Delta}=\sqrt{388}=\sqrt{4*97}=\sqrt{4}*\sqrt{97}=2\sqrt{97}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{97}}{2*1}=\frac{14-2\sqrt{97}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{97}}{2*1}=\frac{14+2\sqrt{97}}{2}
$