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3a^2=17
We move all terms to the left:
3a^2-(17)=0
a = 3; b = 0; c = -17;
Δ = b2-4ac
Δ = 02-4·3·(-17)
Δ = 204
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{204}=\sqrt{4*51}=\sqrt{4}*\sqrt{51}=2\sqrt{51}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{51}}{2*3}=\frac{0-2\sqrt{51}}{6} =-\frac{2\sqrt{51}}{6} =-\frac{\sqrt{51}}{3} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{51}}{2*3}=\frac{0+2\sqrt{51}}{6} =\frac{2\sqrt{51}}{6} =\frac{\sqrt{51}}{3} $
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