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(3x+6)(x+4)=0
We multiply parentheses ..
(+3x^2+12x+6x+24)=0
We get rid of parentheses
3x^2+12x+6x+24=0
We add all the numbers together, and all the variables
3x^2+18x+24=0
a = 3; b = 18; c = +24;
Δ = b2-4ac
Δ = 182-4·3·24
Δ = 36
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{36}=6$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6}{2*3}=\frac{-24}{6} =-4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6}{2*3}=\frac{-12}{6} =-2 $
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