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48=3x^2
We move all terms to the left:
48-(3x^2)=0
a = -3; b = 0; c = +48;
Δ = b2-4ac
Δ = 02-4·(-3)·48
Δ = 576
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{576}=24$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24}{2*-3}=\frac{-24}{-6} =+4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24}{2*-3}=\frac{24}{-6} =-4 $
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