26q=-19q^2

Solution for 26q=-19q^2 equation:

26q=-19q^2

We move all terms to the left:

26q-(-19q^2)=0

We get rid of parentheses

19q^2+26q=0

a = 19; b = 26; c = 0;
Δ = b2-4ac
Δ = 262-4·19·0
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-26}{2*19}=\frac{-52}{38}
=-1+7/19
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+26}{2*19}=\frac{0}{38}
=0
$