x^2+0.9x-0.3=0

Solution for x^2+0.9x-0.3=0 equation:

x^2+0.9x-0.3=0

a = 1; b = 0.9; c = -0.3;
Δ = b2-4ac
Δ = 0.92-4·1·(-0.3)
Δ = 2.01
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{-(0.9)-\sqrt{2.01}}{2*1}=\frac{-0.9-\sqrt{2.01}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.9)+\sqrt{2.01}}{2*1}=\frac{-0.9+\sqrt{2.01}}{2}
$