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39=3x^2+4x
We move all terms to the left:
39-(3x^2+4x)=0
We get rid of parentheses
-3x^2-4x+39=0
a = -3; b = -4; c = +39;
Δ = b2-4ac
Δ = -42-4·(-3)·39
Δ = 484
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{484}=22$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-22}{2*-3}=\frac{-18}{-6} =+3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+22}{2*-3}=\frac{26}{-6} =-4+1/3 $
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