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(w+3)(w+10)=0
We multiply parentheses ..
(+w^2+10w+3w+30)=0
We get rid of parentheses
w^2+10w+3w+30=0
We add all the numbers together, and all the variables
w^2+13w+30=0
a = 1; b = 13; c = +30;
Δ = b2-4ac
Δ = 132-4·1·30
Δ = 49
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{49}=7$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-7}{2*1}=\frac{-20}{2} =-10 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+7}{2*1}=\frac{-6}{2} =-3 $
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