0.4x^2+19x=50

Solution for 0.4x^2+19x=50 equation:

0.4x^2+19x=50

We move all terms to the left:

0.4x^2+19x-(50)=0

a = 0.4; b = 19; c = -50;
Δ = b2-4ac
Δ = 192-4·0.4·(-50)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-21}{2*0.4}=\frac{-40}{0.8}
=-50
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+21}{2*0.4}=\frac{2}{0.8}
=2+0.4/0.8
$