W(x)=-0.01x^2+0.5x+0.06

Solution for W(x)=-0.01x^2+0.5x+0.06 equation:

(W)=-0.01W^2+0.5W+0.06

We move all terms to the left:

(W)-(-0.01W^2+0.5W+0.06)=0

We get rid of parentheses

0.01W^2-0.5W+W-0.06=0

We add all the numbers together, and all the variables

0.01W^2+0.5W-0.06=0

a = 0.01; b = 0.5; c = -0.06;
Δ = b2-4ac
Δ = 0.52-4·0.01·(-0.06)
Δ = 0.2524
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$W_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$W_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$W_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0.5)-\sqrt{0.2524}}{2*0.01}=\frac{-0.5-\sqrt{0.2524}}{0.02}
$
$W_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.5)+\sqrt{0.2524}}{2*0.01}=\frac{-0.5+\sqrt{0.2524}}{0.02}
$