0,1x^2+12x-16000=0

Solution for 0,1x^2+12x-16000=0 equation:

0.1x^2+12x-16000=0

a = 0.1; b = 12; c = -16000;
Δ = b2-4ac
Δ = 122-4·0.1·(-16000)
Δ = 6544
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:

$\sqrt{\Delta}=\sqrt{6544}=\sqrt{16*409}=\sqrt{16}*\sqrt{409}=4\sqrt{409}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{409}}{2*0.1}=\frac{-12-4\sqrt{409}}{0.2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{409}}{2*0.1}=\frac{-12+4\sqrt{409}}{0.2}
$