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2(3)+3x^2=48
We move all terms to the left:
2(3)+3x^2-(48)=0
We add all the numbers together, and all the variables
3x^2-25=0
a = 3; b = 0; c = -25;
Δ = b2-4ac
Δ = 02-4·3·(-25)
Δ = 300
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{300}=\sqrt{100*3}=\sqrt{100}*\sqrt{3}=10\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{3}}{2*3}=\frac{0-10\sqrt{3}}{6} =-\frac{10\sqrt{3}}{6} =-\frac{5\sqrt{3}}{3} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{3}}{2*3}=\frac{0+10\sqrt{3}}{6} =\frac{10\sqrt{3}}{6} =\frac{5\sqrt{3}}{3} $
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