x^2+0.1x-4.62=0

Solution for x^2+0.1x-4.62=0 equation:

x^2+0.1x-4.62=0

a = 1; b = 0.1; c = -4.62;
Δ = b2-4ac
Δ = 0.12-4·1·(-4.62)
Δ = 18.49
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{-(0.1)-\sqrt{18.49}}{2*1}=\frac{-0.1-\sqrt{18.49}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.1)+\sqrt{18.49}}{2*1}=\frac{-0.1+\sqrt{18.49}}{2}
$