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3x^2+4=16
We move all terms to the left:
3x^2+4-(16)=0
We add all the numbers together, and all the variables
3x^2-12=0
a = 3; b = 0; c = -12;
Δ = b2-4ac
Δ = 02-4·3·(-12)
Δ = 144
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{144}=12$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12}{2*3}=\frac{-12}{6} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12}{2*3}=\frac{12}{6} =2 $
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