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