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3x^2=212
We move all terms to the left:
3x^2-(212)=0
a = 3; b = 0; c = -212;
Δ = b2-4ac
Δ = 02-4·3·(-212)
Δ = 2544
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2544}=\sqrt{16*159}=\sqrt{16}*\sqrt{159}=4\sqrt{159}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{159}}{2*3}=\frac{0-4\sqrt{159}}{6} =-\frac{4\sqrt{159}}{6} =-\frac{2\sqrt{159}}{3} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{159}}{2*3}=\frac{0+4\sqrt{159}}{6} =\frac{4\sqrt{159}}{6} =\frac{2\sqrt{159}}{3} $
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