0=0.1x^2+12x+36

Solution for 0=0.1x^2+12x+36 equation:

0=0.1x^2+12x+36

We move all terms to the left:

0-(0.1x^2+12x+36)=0

We add all the numbers together, and all the variables

-(0.1x^2+12x+36)=0

We get rid of parentheses

-0.1x^2-12x-36=0

a = -0.1; b = -12; c = -36;
Δ = b2-4ac
Δ = -122-4·(-0.1)·(-36)
Δ = 129.6
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{-(-12)-\sqrt{129.6}}{2*-0.1}=\frac{12-\sqrt{129.6}}{-0.2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+\sqrt{129.6}}{2*-0.1}=\frac{12+\sqrt{129.6}}{-0.2}
$