115=0.05x^2-4x-38

Solution for 115=0.05x^2-4x-38 equation:

115=0.05x^2-4x-38

We move all terms to the left:

115-(0.05x^2-4x-38)=0

We get rid of parentheses

-0.05x^2+4x+38+115=0

We add all the numbers together, and all the variables

-0.05x^2+4x+153=0

a = -0.05; b = 4; c = +153;
Δ = b2-4ac
Δ = 42-4·(-0.05)·153
Δ = 46.6
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{-(4)-\sqrt{46.6}}{2*-0.05}=\frac{-4-\sqrt{46.6}}{-0.1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+\sqrt{46.6}}{2*-0.05}=\frac{-4+\sqrt{46.6}}{-0.1}
$