16t^2+4t-39=0

Solution for 16t^2+4t-39=0 equation:

16t^2+4t-39=0

a = 16; b = 4; c = -39;
Δ = b2-4ac
Δ = 42-4·16·(-39)
Δ = 2512
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:

$\sqrt{\Delta}=\sqrt{2512}=\sqrt{16*157}=\sqrt{16}*\sqrt{157}=4\sqrt{157}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{157}}{2*16}=\frac{-4-4\sqrt{157}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{157}}{2*16}=\frac{-4+4\sqrt{157}}{32}
$