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21x^2-49x+14=0
a = 21; b = -49; c = +14;
Δ = b2-4ac
Δ = -492-4·21·14
Δ = 1225
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{1225}=35$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-49)-35}{2*21}=\frac{14}{42} =1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-49)+35}{2*21}=\frac{84}{42} =2 $
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