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p^2+4p-51=0
a = 1; b = 4; c = -51;
Δ = b2-4ac
Δ = 42-4·1·(-51)
Δ = 220
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}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{220}=\sqrt{4*55}=\sqrt{4}*\sqrt{55}=2\sqrt{55}$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{55}}{2*1}=\frac{-4-2\sqrt{55}}{2} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{55}}{2*1}=\frac{-4+2\sqrt{55}}{2} $
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