z2+0.3z-0.1=0

Solution for z2+0.3z-0.1=0 equation:

z2+0.3z-0.1=0

We add all the numbers together, and all the variables

z^2+0.3z-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:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0.3)-\sqrt{0.49}}{2*1}=\frac{-0.3-\sqrt{0.49}}{2}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.3)+\sqrt{0.49}}{2*1}=\frac{-0.3+\sqrt{0.49}}{2}
$