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p2+12p-54=0
We add all the numbers together, and all the variables
p^2+12p-54=0
a = 1; b = 12; c = -54;
Δ = b2-4ac
Δ = 122-4·1·(-54)
Δ = 360
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}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{360}=\sqrt{36*10}=\sqrt{36}*\sqrt{10}=6\sqrt{10}$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-6\sqrt{10}}{2*1}=\frac{-12-6\sqrt{10}}{2} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+6\sqrt{10}}{2*1}=\frac{-12+6\sqrt{10}}{2} $
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