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3x^2-294=0
a = 3; b = 0; c = -294;
Δ = b2-4ac
Δ = 02-4·3·(-294)
Δ = 3528
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3528}=\sqrt{1764*2}=\sqrt{1764}*\sqrt{2}=42\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-42\sqrt{2}}{2*3}=\frac{0-42\sqrt{2}}{6} =-\frac{42\sqrt{2}}{6} =-7\sqrt{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+42\sqrt{2}}{2*3}=\frac{0+42\sqrt{2}}{6} =\frac{42\sqrt{2}}{6} =7\sqrt{2} $
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