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2(x+4)3x=17
We move all terms to the left:
2(x+4)3x-(17)=0
We multiply parentheses
6x^2+24x-17=0
a = 6; b = 24; c = -17;
Δ = b2-4ac
Δ = 242-4·6·(-17)
Δ = 984
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{984}=\sqrt{4*246}=\sqrt{4}*\sqrt{246}=2\sqrt{246}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-2\sqrt{246}}{2*6}=\frac{-24-2\sqrt{246}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+2\sqrt{246}}{2*6}=\frac{-24+2\sqrt{246}}{12} $
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