0=-16t^2+212t+139

Solution for 0=-16t^2+212t+139 equation:

0=-16t^2+212t+139

We move all terms to the left:

0-(-16t^2+212t+139)=0

We add all the numbers together, and all the variables

-(-16t^2+212t+139)=0

We get rid of parentheses

16t^2-212t-139=0

a = 16; b = -212; c = -139;
Δ = b2-4ac
Δ = -2122-4·16·(-139)
Δ = 53840
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:

$\sqrt{\Delta}=\sqrt{53840}=\sqrt{16*3365}=\sqrt{16}*\sqrt{3365}=4\sqrt{3365}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-212)-4\sqrt{3365}}{2*16}=\frac{212-4\sqrt{3365}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-212)+4\sqrt{3365}}{2*16}=\frac{212+4\sqrt{3365}}{32}
$