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21x^2-41x+18=0
a = 21; b = -41; c = +18;
Δ = b2-4ac
Δ = -412-4·21·18
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{169}=13$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-41)-13}{2*21}=\frac{28}{42} =2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-41)+13}{2*21}=\frac{54}{42} =1+2/7 $
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