w2+5w-32=2w-4

Solution for w2+5w-32=2w-4 equation:

w2+5w-32=2w-4

We move all terms to the left:

w2+5w-32-(2w-4)=0

We add all the numbers together, and all the variables

w^2+5w-(2w-4)-32=0

We get rid of parentheses

w^2+5w-2w+4-32=0

We add all the numbers together, and all the variables

w^2+3w-28=0

a = 1; b = 3; c = -28;
Δ = b2-4ac
Δ = 32-4·1·(-28)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-11}{2*1}=\frac{-14}{2}
=-7
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+11}{2*1}=\frac{8}{2}
=4
$