0,2x^2+6x-80=0

Solution for 0,2x^2+6x-80=0 equation:

0.2x^2+6x-80=0

a = 0.2; b = 6; c = -80;
Δ = b2-4ac
Δ = 62-4·0.2·(-80)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-10}{2*0.2}=\frac{-16}{0.4}
=-40
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+10}{2*0.2}=\frac{4}{0.4}
=10
$