12w^2+16w=48

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

12w^2+16w=48

We move all terms to the left:

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

a = 12; b = 16; c = -48;
Δ = b2-4ac
Δ = 162-4·12·(-48)
Δ = 2560
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:

$\sqrt{\Delta}=\sqrt{2560}=\sqrt{256*10}=\sqrt{256}*\sqrt{10}=16\sqrt{10}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16\sqrt{10}}{2*12}=\frac{-16-16\sqrt{10}}{24}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16\sqrt{10}}{2*12}=\frac{-16+16\sqrt{10}}{24}
$