(x^2)+0.3x-0.1=0

Solution for (x^2)+0.3x-0.1=0 equation:

(x^2)+0.3x-0.1=0

a = 1; b = 0.3; c = -0.1;
Δ = b2-4ac
Δ = 0.32-4·1·(-0.1)
Δ = 0.49
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{-(0.3)-\sqrt{0.49}}{2*1}=\frac{-0.3-\sqrt{0.49}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.3)+\sqrt{0.49}}{2*1}=\frac{-0.3+\sqrt{0.49}}{2}
$