0=(-16t^2)+64t+80

Solution for 0=(-16t^2)+64t+80 equation:

0=(-16t^2)+64t+80

We move all terms to the left:

0-((-16t^2)+64t+80)=0

We add all the numbers together, and all the variables

-((-16t^2)+64t+80)=0


We calculate terms in parentheses: -((-16t^2)+64t+80), so:

(-16t^2)+64t+80

We get rid of parentheses

-16t^2+64t+80

Back to the equation:

-(-16t^2+64t+80)


We get rid of parentheses

16t^2-64t-80=0

a = 16; b = -64; c = -80;
Δ = b2-4ac
Δ = -642-4·16·(-80)
Δ = 9216
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9216}=96$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-64)-96}{2*16}=\frac{-32}{32}
=-1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-64)+96}{2*16}=\frac{160}{32}
=5
$