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(3x+2)2x=17
We move all terms to the left:
(3x+2)2x-(17)=0
We multiply parentheses
6x^2+4x-17=0
a = 6; b = 4; c = -17;
Δ = b2-4ac
Δ = 42-4·6·(-17)
Δ = 424
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{424}=\sqrt{4*106}=\sqrt{4}*\sqrt{106}=2\sqrt{106}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{106}}{2*6}=\frac{-4-2\sqrt{106}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{106}}{2*6}=\frac{-4+2\sqrt{106}}{12} $
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