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21x^2-31x+4=0
a = 21; b = -31; c = +4;
Δ = b2-4ac
Δ = -312-4·21·4
Δ = 625
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{625}=25$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-31)-25}{2*21}=\frac{6}{42} =1/7 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+25}{2*21}=\frac{56}{42} =1+1/3 $
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