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(w+5)(w+2)=10
We move all terms to the left:
(w+5)(w+2)-(10)=0
We multiply parentheses ..
(+w^2+2w+5w+10)-10=0
We get rid of parentheses
w^2+2w+5w+10-10=0
We add all the numbers together, and all the variables
w^2+7w=0
a = 1; b = 7; c = 0;
Δ = b2-4ac
Δ = 72-4·1·0
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$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-7}{2*1}=\frac{-14}{2} =-7 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+7}{2*1}=\frac{0}{2} =0 $
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