.1x^2+12x-100=10

Solution for .1x^2+12x-100=10 equation:

.1x^2+12x-100=10

We move all terms to the left:

.1x^2+12x-100-(10)=0

We add all the numbers together, and all the variables

.1x^2+12x-110=0

a = .1; b = 12; c = -110;
Δ = b2-4ac
Δ = 122-4·.1·(-110)
Δ = 188
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:

$\sqrt{\Delta}=\sqrt{188}=\sqrt{4*47}=\sqrt{4}*\sqrt{47}=2\sqrt{47}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{47}}{2*.1}=\frac{-12-2\sqrt{47}}{0.2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{47}}{2*.1}=\frac{-12+2\sqrt{47}}{0.2}
$