1,4q+q^2=0,51

Solution for 1,4q+q^2=0,51 equation:

1.4q+q^2=0.51

We move all terms to the left:

1.4q+q^2-(0.51)=0

We add all the numbers together, and all the variables

q^2+1.4q-0.51=0

a = 1; b = 1.4; c = -0.51;
Δ = b2-4ac
Δ = 1.42-4·1·(-0.51)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1.4)-2}{2*1}=\frac{-3.4}{2}
=-1+1/1.4285714285714
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.4)+2}{2*1}=\frac{0.6}{2}
=0.6/2
$