0.2x^2+6x+5=0

Solution for 0.2x^2+6x+5=0 equation:

0.2x^2+6x+5=0

a = 0.2; b = 6; c = +5;
Δ = b2-4ac
Δ = 62-4·0.2·5
Δ = 32
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-4\sqrt{2}}{2*0.2}=\frac{-6-4\sqrt{2}}{0.4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+4\sqrt{2}}{2*0.2}=\frac{-6+4\sqrt{2}}{0.4}
$