9t^2+48t-100=0

Solution for 9t^2+48t-100=0 equation:

9t^2+48t-100=0

a = 9; b = 48; c = -100;
Δ = b2-4ac
Δ = 482-4·9·(-100)
Δ = 5904
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:

$\sqrt{\Delta}=\sqrt{5904}=\sqrt{144*41}=\sqrt{144}*\sqrt{41}=12\sqrt{41}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-12\sqrt{41}}{2*9}=\frac{-48-12\sqrt{41}}{18}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+12\sqrt{41}}{2*9}=\frac{-48+12\sqrt{41}}{18}
$