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21x^2-14x+2=0
a = 21; b = -14; c = +2;
Δ = b2-4ac
Δ = -142-4·21·2
Δ = 28
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{7}}{2*21}=\frac{14-2\sqrt{7}}{42} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{7}}{2*21}=\frac{14+2\sqrt{7}}{42} $
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