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2w^2-15w+28=0
a = 2; b = -15; c = +28;
Δ = b2-4ac
Δ = -152-4·2·28
Δ = 1
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{1}=1$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-1}{2*2}=\frac{14}{4} =3+1/2 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+1}{2*2}=\frac{16}{4} =4 $
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