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28b^2-53b+24=0
a = 28; b = -53; c = +24;
Δ = b2-4ac
Δ = -532-4·28·24
Δ = 121
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{121}=11$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-53)-11}{2*28}=\frac{42}{56} =3/4 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-53)+11}{2*28}=\frac{64}{56} =1+1/7 $
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