-0.2x^2+x+1.4=0

Solution for -0.2x^2+x+1.4=0 equation:

-0.2x^2+x+1.4=0

a = -0.2; b = 1; c = +1.4;
Δ = b2-4ac
Δ = 12-4·(-0.2)·1.4
Δ = 2.12
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)-\sqrt{2.12}}{2*-0.2}=\frac{-1-\sqrt{2.12}}{-0.4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{2.12}}{2*-0.2}=\frac{-1+\sqrt{2.12}}{-0.4}
$