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21p^2+10p-1=0
a = 21; b = 10; c = -1;
Δ = b2-4ac
Δ = 102-4·21·(-1)
Δ = 184
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{184}=\sqrt{4*46}=\sqrt{4}*\sqrt{46}=2\sqrt{46}$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{46}}{2*21}=\frac{-10-2\sqrt{46}}{42} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{46}}{2*21}=\frac{-10+2\sqrt{46}}{42} $
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