16t^2+16t-48=0

Solution for 16t^2+16t-48=0 equation:

16t^2+16t-48=0

a = 16; b = 16; c = -48;
Δ = b2-4ac
Δ = 162-4·16·(-48)
Δ = 3328
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:

$\sqrt{\Delta}=\sqrt{3328}=\sqrt{256*13}=\sqrt{256}*\sqrt{13}=16\sqrt{13}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16\sqrt{13}}{2*16}=\frac{-16-16\sqrt{13}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16\sqrt{13}}{2*16}=\frac{-16+16\sqrt{13}}{32}
$