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5w^2-31w+28=0
a = 5; b = -31; c = +28;
Δ = b2-4ac
Δ = -312-4·5·28
Δ = 401
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}$$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-31)-\sqrt{401}}{2*5}=\frac{31-\sqrt{401}}{10} $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+\sqrt{401}}{2*5}=\frac{31+\sqrt{401}}{10} $
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