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3x^2+2=13
We move all terms to the left:
3x^2+2-(13)=0
We add all the numbers together, and all the variables
3x^2-11=0
a = 3; b = 0; c = -11;
Δ = b2-4ac
Δ = 02-4·3·(-11)
Δ = 132
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{33}}{2*3}=\frac{0-2\sqrt{33}}{6} =-\frac{2\sqrt{33}}{6} =-\frac{\sqrt{33}}{3} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{33}}{2*3}=\frac{0+2\sqrt{33}}{6} =\frac{2\sqrt{33}}{6} =\frac{\sqrt{33}}{3} $
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