40=560-0.2x^2

Solution for 40=560-0.2x^2 equation:

40=560-0.2x^2

We move all terms to the left:

40-(560-0.2x^2)=0

We get rid of parentheses

0.2x^2-560+40=0

We add all the numbers together, and all the variables

0.2x^2-520=0

a = 0.2; b = 0; c = -520;
Δ = b2-4ac
Δ = 02-4·0.2·(-520)
Δ = 416
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{416}=\sqrt{16*26}=\sqrt{16}*\sqrt{26}=4\sqrt{26}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{26}}{2*0.2}=\frac{0-4\sqrt{26}}{0.4}
=-\frac{4\sqrt{26}}{0.4}
=-\frac{\sqrt{26}}{0.1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{26}}{2*0.2}=\frac{0+4\sqrt{26}}{0.4}
=\frac{4\sqrt{26}}{0.4}
=\frac{\sqrt{26}}{0.1}
$