A(t)=-3t2+10t+80

Solution for A(t)=-3t2+10t+80 equation:

(A)=-3A^2+10A+80

We move all terms to the left:

(A)-(-3A^2+10A+80)=0

We get rid of parentheses

3A^2-10A+A-80=0

We add all the numbers together, and all the variables

3A^2-9A-80=0

a = 3; b = -9; c = -80;
Δ = b2-4ac
Δ = -92-4·3·(-80)
Δ = 1041
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-\sqrt{1041}}{2*3}=\frac{9-\sqrt{1041}}{6}
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{1041}}{2*3}=\frac{9+\sqrt{1041}}{6}
$