16t^2+138t-78=0

Solution for 16t^2+138t-78=0 equation:

16t^2+138t-78=0

a = 16; b = 138; c = -78;
Δ = b2-4ac
Δ = 1382-4·16·(-78)
Δ = 24036
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:

$\sqrt{\Delta}=\sqrt{24036}=\sqrt{4*6009}=\sqrt{4}*\sqrt{6009}=2\sqrt{6009}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(138)-2\sqrt{6009}}{2*16}=\frac{-138-2\sqrt{6009}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(138)+2\sqrt{6009}}{2*16}=\frac{-138+2\sqrt{6009}}{32}
$