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(x+3)(6x+1)=0
We multiply parentheses ..
(+6x^2+x+18x+3)=0
We get rid of parentheses
6x^2+x+18x+3=0
We add all the numbers together, and all the variables
6x^2+19x+3=0
a = 6; b = 19; c = +3;
Δ = b2-4ac
Δ = 192-4·6·3
Δ = 289
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{289}=17$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-17}{2*6}=\frac{-36}{12} =-3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+17}{2*6}=\frac{-2}{12} =-1/6 $
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