40-4x=10+0.4x^2

Solution for 40-4x=10+0.4x^2 equation:

40-4x=10+0.4x^2

We move all terms to the left:

40-4x-(10+0.4x^2)=0

We get rid of parentheses

-0.4x^2-4x-10+40=0

We add all the numbers together, and all the variables

-0.4x^2-4x+30=0

a = -0.4; b = -4; c = +30;
Δ = b2-4ac
Δ = -42-4·(-0.4)·30
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8}{2*-0.4}=\frac{-4}{-0.8}
=+5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8}{2*-0.4}=\frac{12}{-0.8}
=-15
$