0.3x^2+1.5x+0.8=0

Solution for 0.3x^2+1.5x+0.8=0 equation:

0.3x^2+1.5x+0.8=0

a = 0.3; b = 1.5; c = +0.8;
Δ = b2-4ac
Δ = 1.52-4·0.3·0.8
Δ = 1.29
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{-(1.5)-\sqrt{1.29}}{2*0.3}=\frac{-1.5-\sqrt{1.29}}{0.6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.5)+\sqrt{1.29}}{2*0.3}=\frac{-1.5+\sqrt{1.29}}{0.6}
$