X^2+0.6x+0.01=0

Solution for X^2+0.6x+0.01=0 equation:

X^2+0.6X+0.01=0

a = 1; b = 0.6; c = +0.01;
Δ = b2-4ac
Δ = 0.62-4·1·0.01
Δ = 0.32
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{0.32}}{2*1}=\frac{-0.6-\sqrt{0.32}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.6)+\sqrt{0.32}}{2*1}=\frac{-0.6+\sqrt{0.32}}{2}
$