t=4t^2+80t

Solution for t=4t^2+80t equation:

t=4t^2+80t

We move all terms to the left:

t-(4t^2+80t)=0

We get rid of parentheses

-4t^2+t-80t=0

We add all the numbers together, and all the variables

-4t^2-79t=0

a = -4; b = -79; c = 0;
Δ = b2-4ac
Δ = -792-4·(-4)·0
Δ = 6241
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6241}=79$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-79)-79}{2*-4}=\frac{0}{-8}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-79)+79}{2*-4}=\frac{158}{-8}
=-19+3/4
$