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-p^2+2p+9=0
We add all the numbers together, and all the variables
-1p^2+2p+9=0
a = -1; b = 2; c = +9;
Δ = b2-4ac
Δ = 22-4·(-1)·9
Δ = 40
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}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{10}}{2*-1}=\frac{-2-2\sqrt{10}}{-2} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{10}}{2*-1}=\frac{-2+2\sqrt{10}}{-2} $
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