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6x^2+4x=9
We move all terms to the left:
6x^2+4x-(9)=0
a = 6; b = 4; c = -9;
Δ = b2-4ac
Δ = 42-4·6·(-9)
Δ = 232
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{232}=\sqrt{4*58}=\sqrt{4}*\sqrt{58}=2\sqrt{58}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{58}}{2*6}=\frac{-4-2\sqrt{58}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{58}}{2*6}=\frac{-4+2\sqrt{58}}{12} $
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