A=12+18t-4.9t^2

Solution for A=12+18t-4.9t^2 equation:

=12+18A-4.9A^2

We move all terms to the left:

-(12+18A-4.9A^2)=0

We get rid of parentheses

4.9A^2-18A-12=0

a = 4.9; b = -18; c = -12;
Δ = b2-4ac
Δ = -182-4·4.9·(-12)
Δ = 559.2
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{-(-18)-\sqrt{559.2}}{2*4.9}=\frac{18-\sqrt{559.2}}{9.8}
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+\sqrt{559.2}}{2*4.9}=\frac{18+\sqrt{559.2}}{9.8}
$