(2w+1)(6+w)=0

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Solution for (2w+1)(6+w)=0 equation:



(2w+1)(6+w)=0
We add all the numbers together, and all the variables
(2w+1)(w+6)=0
We multiply parentheses ..
(+2w^2+12w+w+6)=0
We get rid of parentheses
2w^2+12w+w+6=0
We add all the numbers together, and all the variables
2w^2+13w+6=0
a = 2; b = 13; c = +6;
Δ = b2-4ac
Δ = 132-4·2·6
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-11}{2*2}=\frac{-24}{4} =-6 $
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+11}{2*2}=\frac{-2}{4} =-1/2 $

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