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6x^2-3=11
We move all terms to the left:
6x^2-3-(11)=0
We add all the numbers together, and all the variables
6x^2-14=0
a = 6; b = 0; c = -14;
Δ = b2-4ac
Δ = 02-4·6·(-14)
Δ = 336
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{336}=\sqrt{16*21}=\sqrt{16}*\sqrt{21}=4\sqrt{21}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{21}}{2*6}=\frac{0-4\sqrt{21}}{12} =-\frac{4\sqrt{21}}{12} =-\frac{\sqrt{21}}{3} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{21}}{2*6}=\frac{0+4\sqrt{21}}{12} =\frac{4\sqrt{21}}{12} =\frac{\sqrt{21}}{3} $
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