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(w+5)(w+7)=15
We move all terms to the left:
(w+5)(w+7)-(15)=0
We multiply parentheses ..
(+w^2+7w+5w+35)-15=0
We get rid of parentheses
w^2+7w+5w+35-15=0
We add all the numbers together, and all the variables
w^2+12w+20=0
a = 1; b = 12; c = +20;
Δ = b2-4ac
Δ = 122-4·1·20
Δ = 64
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{64}=8$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8}{2*1}=\frac{-20}{2} =-10 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8}{2*1}=\frac{-4}{2} =-2 $
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