q^2+26q+9=-43

Solution for q^2+26q+9=-43 equation:

q^2+26q+9=-43

We move all terms to the left:

q^2+26q+9-(-43)=0

We add all the numbers together, and all the variables

q^2+26q+52=0

a = 1; b = 26; c = +52;
Δ = b2-4ac
Δ = 262-4·1·52
Δ = 468
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{468}=\sqrt{36*13}=\sqrt{36}*\sqrt{13}=6\sqrt{13}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-6\sqrt{13}}{2*1}=\frac{-26-6\sqrt{13}}{2}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+6\sqrt{13}}{2*1}=\frac{-26+6\sqrt{13}}{2}
$