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(3x+1)(2x=4)
We move all terms to the left:
(3x+1)(2x-(4))=0
We multiply parentheses ..
(+6x^2-12x+2x-4)=0
We get rid of parentheses
6x^2-12x+2x-4=0
We add all the numbers together, and all the variables
6x^2-10x-4=0
a = 6; b = -10; c = -4;
Δ = b2-4ac
Δ = -102-4·6·(-4)
Δ = 196
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{196}=14$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-14}{2*6}=\frac{-4}{12} =-1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+14}{2*6}=\frac{24}{12} =2 $
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