A(t)=-t(2)+6t+18

Solution for A(t)=-t(2)+6t+18 equation:

(A)=-A(2)+6A+18

We move all terms to the left:

(A)-(-A(2)+6A+18)=0

We add all the numbers together, and all the variables

-(-1A^2+6A+18)+A=0

We get rid of parentheses

1A^2-6A+A-18=0

We add all the numbers together, and all the variables

A^2-5A-18=0

a = 1; b = -5; c = -18;
Δ = b2-4ac
Δ = -52-4·1·(-18)
Δ = 97
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{-(-5)-\sqrt{97}}{2*1}=\frac{5-\sqrt{97}}{2}
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{97}}{2*1}=\frac{5+\sqrt{97}}{2}
$