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10d^2-41d+40=0
a = 10; b = -41; c = +40;
Δ = b2-4ac
Δ = -412-4·10·40
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{81}=9$$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-41)-9}{2*10}=\frac{32}{20} =1+3/5 $$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-41)+9}{2*10}=\frac{50}{20} =2+1/2 $
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