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2x(3x+4)=14
We move all terms to the left:
2x(3x+4)-(14)=0
We multiply parentheses
6x^2+8x-14=0
a = 6; b = 8; c = -14;
Δ = b2-4ac
Δ = 82-4·6·(-14)
Δ = 400
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{400}=20$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-20}{2*6}=\frac{-28}{12} =-2+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+20}{2*6}=\frac{12}{12} =1 $
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