2w+w^2=48

Solution for 2w+w^2=48 equation:

2w+w^2=48

We move all terms to the left:

2w+w^2-(48)=0

a = 1; b = 2; c = -48;
Δ = b2-4ac
Δ = 22-4·1·(-48)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-14}{2*1}=\frac{-16}{2}
=-8
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+14}{2*1}=\frac{12}{2}
=6
$