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10p^2-20p+3=0
a = 10; b = -20; c = +3;
Δ = b2-4ac
Δ = -202-4·10·3
Δ = 280
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}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{280}=\sqrt{4*70}=\sqrt{4}*\sqrt{70}=2\sqrt{70}$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{70}}{2*10}=\frac{20-2\sqrt{70}}{20} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{70}}{2*10}=\frac{20+2\sqrt{70}}{20} $
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