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28b^2-96b+80=0
a = 28; b = -96; c = +80;
Δ = b2-4ac
Δ = -962-4·28·80
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{256}=16$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-96)-16}{2*28}=\frac{80}{56} =1+3/7 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-96)+16}{2*28}=\frac{112}{56} =2 $
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