15t^2-96t+48=0

Solution for 15t^2-96t+48=0 equation:

15t^2-96t+48=0

a = 15; b = -96; c = +48;
Δ = b2-4ac
Δ = -962-4·15·48
Δ = 6336
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:

$\sqrt{\Delta}=\sqrt{6336}=\sqrt{576*11}=\sqrt{576}*\sqrt{11}=24\sqrt{11}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-96)-24\sqrt{11}}{2*15}=\frac{96-24\sqrt{11}}{30}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-96)+24\sqrt{11}}{2*15}=\frac{96+24\sqrt{11}}{30}
$