x^2+0.2x=50

Solution for x^2+0.2x=50 equation:

x^2+0.2x=50

We move all terms to the left:

x^2+0.2x-(50)=0

a = 1; b = 0.2; c = -50;
Δ = b2-4ac
Δ = 0.22-4·1·(-50)
Δ = 200.04
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{-(0.2)-\sqrt{200.04}}{2*1}=\frac{-0.2-\sqrt{200.04}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.2)+\sqrt{200.04}}{2*1}=\frac{-0.2+\sqrt{200.04}}{2}
$