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(5+w)(4w+3)=0
We add all the numbers together, and all the variables
(w+5)(4w+3)=0
We multiply parentheses ..
(+4w^2+3w+20w+15)=0
We get rid of parentheses
4w^2+3w+20w+15=0
We add all the numbers together, and all the variables
4w^2+23w+15=0
a = 4; b = 23; c = +15;
Δ = b2-4ac
Δ = 232-4·4·15
Δ = 289
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{289}=17$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-17}{2*4}=\frac{-40}{8} =-5 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+17}{2*4}=\frac{-6}{8} =-3/4 $
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