q^2+4q=47

Solution for q^2+4q=47 equation:

q^2+4q=47

We move all terms to the left:

q^2+4q-(47)=0

a = 1; b = 4; c = -47;
Δ = b2-4ac
Δ = 42-4·1·(-47)
Δ = 204
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:

$\sqrt{\Delta}=\sqrt{204}=\sqrt{4*51}=\sqrt{4}*\sqrt{51}=2\sqrt{51}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{51}}{2*1}=\frac{-4-2\sqrt{51}}{2}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{51}}{2*1}=\frac{-4+2\sqrt{51}}{2}
$