A(t)=4t+3t^2

Solution for A(t)=4t+3t^2 equation:

(A)=4A+3A^2

We move all terms to the left:

(A)-(4A+3A^2)=0

We get rid of parentheses

-3A^2-4A+A=0

We add all the numbers together, and all the variables

-3A^2-3A=0

a = -3; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·(-3)·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*-3}=\frac{0}{-6}
=0
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*-3}=\frac{6}{-6}
=-1
$