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(3x+1)(x-2)=0.
We multiply parentheses ..
(+3x^2-6x+x-2)=0
We get rid of parentheses
3x^2-6x+x-2=0
We add all the numbers together, and all the variables
3x^2-5x-2=0
a = 3; b = -5; c = -2;
Δ = b2-4ac
Δ = -52-4·3·(-2)
Δ = 49
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{49}=7$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-7}{2*3}=\frac{-2}{6} =-1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+7}{2*3}=\frac{12}{6} =2 $
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