x^2+0.6x-0.045=0

Solution for x^2+0.6x-0.045=0 equation:

x^2+0.6x-0.045=0

a = 1; b = 0.6; c = -0.045;
Δ = b2-4ac
Δ = 0.62-4·1·(-0.045)
Δ = 0.54
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{0.54}}{2*1}=\frac{-0.6-\sqrt{0.54}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.6)+\sqrt{0.54}}{2*1}=\frac{-0.6+\sqrt{0.54}}{2}
$