40-x+0.01x^2=40-2x+0.03x^2

Solution for 40-x+0.01x^2=40-2x+0.03x^2 equation:

40-x+0.01x^2=40-2x+0.03x^2

We move all terms to the left:

40-x+0.01x^2-(40-2x+0.03x^2)=0

We add all the numbers together, and all the variables

0.01x^2-(40-2x+0.03x^2)-1x+40=0

We get rid of parentheses

0.01x^2-0.03x^2+2x-1x-40+40=0

We add all the numbers together, and all the variables

-0.02x^2+x=0

a = -0.02; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-0.02)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-0.02}=\frac{-2}{-0.04}
=+50
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-0.02}=\frac{0}{-0.04}
=0
$