x=0.05x^2-4x-38

Solution for x=0.05x^2-4x-38 equation:

x=0.05x^2-4x-38

We move all terms to the left:

x-(0.05x^2-4x-38)=0

We get rid of parentheses

-0.05x^2+x+4x+38=0

We add all the numbers together, and all the variables

-0.05x^2+5x+38=0

a = -0.05; b = 5; c = +38;
Δ = b2-4ac
Δ = 52-4·(-0.05)·38
Δ = 32.6
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{32.6}}{2*-0.05}=\frac{-5-\sqrt{32.6}}{-0.1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{32.6}}{2*-0.05}=\frac{-5+\sqrt{32.6}}{-0.1}
$