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4p^2-30p+11=0
a = 4; b = -30; c = +11;
Δ = b2-4ac
Δ = -302-4·4·11
Δ = 724
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}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{724}=\sqrt{4*181}=\sqrt{4}*\sqrt{181}=2\sqrt{181}$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-2\sqrt{181}}{2*4}=\frac{30-2\sqrt{181}}{8} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+2\sqrt{181}}{2*4}=\frac{30+2\sqrt{181}}{8} $
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