0.1x^2+0.6x+1=10

Solution for 0.1x^2+0.6x+1=10 equation:

0.1x^2+0.6x+1=10

We move all terms to the left:

0.1x^2+0.6x+1-(10)=0

We add all the numbers together, and all the variables

0.1x^2+0.6x-9=0

a = 0.1; b = 0.6; c = -9;
Δ = b2-4ac
Δ = 0.62-4·0.1·(-9)
Δ = 3.96
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{-(0.6)-\sqrt{3.96}}{2*0.1}=\frac{-0.6-\sqrt{3.96}}{0.2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.6)+\sqrt{3.96}}{2*0.1}=\frac{-0.6+\sqrt{3.96}}{0.2}
$