X^2+0.3x-1.2=0

Solution for X^2+0.3x-1.2=0 equation:

X^2+0.3X-1.2=0

a = 1; b = 0.3; c = -1.2;
Δ = b2-4ac
Δ = 0.32-4·1·(-1.2)
Δ = 4.89
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{-(0.3)-\sqrt{4.89}}{2*1}=\frac{-0.3-\sqrt{4.89}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.3)+\sqrt{4.89}}{2*1}=\frac{-0.3+\sqrt{4.89}}{2}
$