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4(3x^2)=14
We move all terms to the left:
4(3x^2)-(14)=0
a = 43; b = 0; c = -14;
Δ = b2-4ac
Δ = 02-4·43·(-14)
Δ = 2408
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2408}=\sqrt{4*602}=\sqrt{4}*\sqrt{602}=2\sqrt{602}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{602}}{2*43}=\frac{0-2\sqrt{602}}{86} =-\frac{2\sqrt{602}}{86} =-\frac{\sqrt{602}}{43} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{602}}{2*43}=\frac{0+2\sqrt{602}}{86} =\frac{2\sqrt{602}}{86} =\frac{\sqrt{602}}{43} $
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