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3p^2-10p+6=0
a = 3; b = -10; c = +6;
Δ = b2-4ac
Δ = -102-4·3·6
Δ = 28
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}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{7}}{2*3}=\frac{10-2\sqrt{7}}{6} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{7}}{2*3}=\frac{10+2\sqrt{7}}{6} $
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