0=w^2-4w-48

Solution for 0=w^2-4w-48 equation:

0=w^2-4w-48

We move all terms to the left:

0-(w^2-4w-48)=0

We add all the numbers together, and all the variables

-(w^2-4w-48)=0

We get rid of parentheses

-w^2+4w+48=0

We add all the numbers together, and all the variables

-1w^2+4w+48=0

a = -1; b = 4; c = +48;
Δ = b2-4ac
Δ = 42-4·(-1)·48
Δ = 208
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:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{13}}{2*-1}=\frac{-4-4\sqrt{13}}{-2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{13}}{2*-1}=\frac{-4+4\sqrt{13}}{-2}
$