2w^2-16w-12w+48=0

Solution for 2w^2-16w-12w+48=0 equation:

2w^2-16w-12w+48=0

We add all the numbers together, and all the variables

2w^2-28w+48=0

a = 2; b = -28; c = +48;
Δ = b2-4ac
Δ = -282-4·2·48
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-20}{2*2}=\frac{8}{4}
=2
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+20}{2*2}=\frac{48}{4}
=12
$