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3x^2=138
We move all terms to the left:
3x^2-(138)=0
a = 3; b = 0; c = -138;
Δ = b2-4ac
Δ = 02-4·3·(-138)
Δ = 1656
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1656}=\sqrt{36*46}=\sqrt{36}*\sqrt{46}=6\sqrt{46}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{46}}{2*3}=\frac{0-6\sqrt{46}}{6} =-\frac{6\sqrt{46}}{6} =-\sqrt{46} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{46}}{2*3}=\frac{0+6\sqrt{46}}{6} =\frac{6\sqrt{46}}{6} =\sqrt{46} $
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