w^2-4w=13

Solution for w^2-4w=13 equation:

w^2-4w=13

We move all terms to the left:

w^2-4w-(13)=0

a = 1; b = -4; c = -13;
Δ = b2-4ac
Δ = -42-4·1·(-13)
Δ = 68
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{17}}{2*1}=\frac{4-2\sqrt{17}}{2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{17}}{2*1}=\frac{4+2\sqrt{17}}{2}
$