16t^2+32t-147=0

Solution for 16t^2+32t-147=0 equation:

16t^2+32t-147=0

a = 16; b = 32; c = -147;
Δ = b2-4ac
Δ = 322-4·16·(-147)
Δ = 10432
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:

$\sqrt{\Delta}=\sqrt{10432}=\sqrt{64*163}=\sqrt{64}*\sqrt{163}=8\sqrt{163}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-8\sqrt{163}}{2*16}=\frac{-32-8\sqrt{163}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+8\sqrt{163}}{2*16}=\frac{-32+8\sqrt{163}}{32}
$