A(w)=-w^2+100

Solution for A(w)=-w^2+100 equation:

(A)=-A^2+100

We move all terms to the left:

(A)-(-A^2+100)=0

We get rid of parentheses

A^2+A-100=0

a = 1; b = 1; c = -100;
Δ = b2-4ac
Δ = 12-4·1·(-100)
Δ = 401
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{-(1)-\sqrt{401}}{2*1}=\frac{-1-\sqrt{401}}{2}
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{401}}{2*1}=\frac{-1+\sqrt{401}}{2}
$