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2w^2+-5w=3
We move all terms to the left:
2w^2+-5w-(3)=0
determiningTheFunctionDomain 2w^2-5w-3+=0
We add all the numbers together, and all the variables
2w^2-5w=0
a = 2; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·2·0
Δ = 25
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{25}=5$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*2}=\frac{0}{4} =0 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*2}=\frac{10}{4} =2+1/2 $
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