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2x(3x+2)=6x+4
We move all terms to the left:
2x(3x+2)-(6x+4)=0
We multiply parentheses
6x^2+4x-(6x+4)=0
We get rid of parentheses
6x^2+4x-6x-4=0
We add all the numbers together, and all the variables
6x^2-2x-4=0
a = 6; b = -2; c = -4;
Δ = b2-4ac
Δ = -22-4·6·(-4)
Δ = 100
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{100}=10$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-10}{2*6}=\frac{-8}{12} =-2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+10}{2*6}=\frac{12}{12} =1 $
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