78=138t-16t^2

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

78=138t-16t^2

We move all terms to the left:

78-(138t-16t^2)=0

We get rid of parentheses

16t^2-138t+78=0

a = 16; b = -138; c = +78;
Δ = b2-4ac
Δ = -1382-4·16·78
Δ = 14052
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:

$\sqrt{\Delta}=\sqrt{14052}=\sqrt{4*3513}=\sqrt{4}*\sqrt{3513}=2\sqrt{3513}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-138)-2\sqrt{3513}}{2*16}=\frac{138-2\sqrt{3513}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-138)+2\sqrt{3513}}{2*16}=\frac{138+2\sqrt{3513}}{32}
$