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(2w+11)(w-5)=0
We multiply parentheses ..
(+2w^2-10w+11w-55)=0
We get rid of parentheses
2w^2-10w+11w-55=0
We add all the numbers together, and all the variables
2w^2+w-55=0
a = 2; b = 1; c = -55;
Δ = b2-4ac
Δ = 12-4·2·(-55)
Δ = 441
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{441}=21$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-21}{2*2}=\frac{-22}{4} =-5+1/2 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+21}{2*2}=\frac{20}{4} =5 $
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