A(w)=-2w^2+300w

Solution for A(w)=-2w^2+300w equation:

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

We move all terms to the left:

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

We get rid of parentheses

2A^2-300A+A=0

We add all the numbers together, and all the variables

2A^2-299A=0

a = 2; b = -299; c = 0;
Δ = b2-4ac
Δ = -2992-4·2·0
Δ = 89401
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{89401}=299$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-299)-299}{2*2}=\frac{0}{4}
=0
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-299)+299}{2*2}=\frac{598}{4}
=149+1/2
$