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3a^2-68=0
a = 3; b = 0; c = -68;
Δ = b2-4ac
Δ = 02-4·3·(-68)
Δ = 816
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}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{816}=\sqrt{16*51}=\sqrt{16}*\sqrt{51}=4\sqrt{51}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{51}}{2*3}=\frac{0-4\sqrt{51}}{6} =-\frac{4\sqrt{51}}{6} =-\frac{2\sqrt{51}}{3} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{51}}{2*3}=\frac{0+4\sqrt{51}}{6} =\frac{4\sqrt{51}}{6} =\frac{2\sqrt{51}}{3} $
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