w^2=(27+3)(27)

Solution for w^2=(27+3)(27) equation:

w^2=(27+3)(27)

We move all terms to the left:

w^2-((27+3)(27))=0

We add all the numbers together, and all the variables

w^2-(3027)=0

We add all the numbers together, and all the variables

w^2-3027=0

a = 1; b = 0; c = -3027;
Δ = b2-4ac
Δ = 02-4·1·(-3027)
Δ = 12108
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:

$\sqrt{\Delta}=\sqrt{12108}=\sqrt{4*3027}=\sqrt{4}*\sqrt{3027}=2\sqrt{3027}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{3027}}{2*1}=\frac{0-2\sqrt{3027}}{2}
=-\frac{2\sqrt{3027}}{2}
=-\sqrt{3027}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{3027}}{2*1}=\frac{0+2\sqrt{3027}}{2}
=\frac{2\sqrt{3027}}{2}
=\sqrt{3027}
$