56=15+0.1x+0.0003x^2

Solution for 56=15+0.1x+0.0003x^2 equation:

56=15+0.1x+0.0003x^2

We move all terms to the left:

56-(15+0.1x+0.0003x^2)=0

We get rid of parentheses

-0.0003x^2-0.1x-15+56=0

We add all the numbers together, and all the variables

-0.0003x^2-0.1x+41=0

a = -0.0003; b = -0.1; c = +41;
Δ = b2-4ac
Δ = -0.12-4·(-0.0003)·41
Δ = 0.0592
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{-(-0.1)-\sqrt{0.0592}}{2*-0.0003}=\frac{0.1-\sqrt{0.0592}}{-0.0006}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.1)+\sqrt{0.0592}}{2*-0.0003}=\frac{0.1+\sqrt{0.0592}}{-0.0006}
$