0=100+12x-0.3x^2

Solution for 0=100+12x-0.3x^2 equation:

0=100+12x-0.3x^2

We move all terms to the left:

0-(100+12x-0.3x^2)=0

We add all the numbers together, and all the variables

-(100+12x-0.3x^2)=0

We get rid of parentheses

0.3x^2-12x-100=0

a = 0.3; b = -12; c = -100;
Δ = b2-4ac
Δ = -122-4·0.3·(-100)
Δ = 264
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{264}=\sqrt{4*66}=\sqrt{4}*\sqrt{66}=2\sqrt{66}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{66}}{2*0.3}=\frac{12-2\sqrt{66}}{0.6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{66}}{2*0.3}=\frac{12+2\sqrt{66}}{0.6}
$