A(t)=40t^2-300t+500

Solution for A(t)=40t^2-300t+500 equation:

(A)=40A^2-300A+500

We move all terms to the left:

(A)-(40A^2-300A+500)=0

We get rid of parentheses

-40A^2+A+300A-500=0

We add all the numbers together, and all the variables

-40A^2+301A-500=0

a = -40; b = 301; c = -500;
Δ = b2-4ac
Δ = 3012-4·(-40)·(-500)
Δ = 10601
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{-(301)-\sqrt{10601}}{2*-40}=\frac{-301-\sqrt{10601}}{-80}
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(301)+\sqrt{10601}}{2*-40}=\frac{-301+\sqrt{10601}}{-80}
$