q^2+26q=-33

Solution for q^2+26q=-33 equation:

q^2+26q=-33

We move all terms to the left:

q^2+26q-(-33)=0

We add all the numbers together, and all the variables

q^2+26q+33=0

a = 1; b = 26; c = +33;
Δ = b2-4ac
Δ = 262-4·1·33
Δ = 544
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:

$\sqrt{\Delta}=\sqrt{544}=\sqrt{16*34}=\sqrt{16}*\sqrt{34}=4\sqrt{34}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-4\sqrt{34}}{2*1}=\frac{-26-4\sqrt{34}}{2}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+4\sqrt{34}}{2*1}=\frac{-26+4\sqrt{34}}{2}
$