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(3w)2=126
We move all terms to the left:
(3w)2-(126)=0
We add all the numbers together, and all the variables
3w^2-126=0
a = 3; b = 0; c = -126;
Δ = b2-4ac
Δ = 02-4·3·(-126)
Δ = 1512
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}$$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1512}=\sqrt{36*42}=\sqrt{36}*\sqrt{42}=6\sqrt{42}$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{42}}{2*3}=\frac{0-6\sqrt{42}}{6} =-\frac{6\sqrt{42}}{6} =-\sqrt{42} $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{42}}{2*3}=\frac{0+6\sqrt{42}}{6} =\frac{6\sqrt{42}}{6} =\sqrt{42} $
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