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21x^2-49x+7=0
a = 21; b = -49; c = +7;
Δ = b2-4ac
Δ = -492-4·21·7
Δ = 1813
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1813}=\sqrt{49*37}=\sqrt{49}*\sqrt{37}=7\sqrt{37}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-49)-7\sqrt{37}}{2*21}=\frac{49-7\sqrt{37}}{42} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-49)+7\sqrt{37}}{2*21}=\frac{49+7\sqrt{37}}{42} $
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