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3b^2-11b+8=0
a = 3; b = -11; c = +8;
Δ = b2-4ac
Δ = -112-4·3·8
Δ = 25
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{25}=5$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-5}{2*3}=\frac{6}{6} =1 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+5}{2*3}=\frac{16}{6} =2+2/3 $
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