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p^2-11p+19=0
a = 1; b = -11; c = +19;
Δ = b2-4ac
Δ = -112-4·1·19
Δ = 45
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}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{45}=\sqrt{9*5}=\sqrt{9}*\sqrt{5}=3\sqrt{5}$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-3\sqrt{5}}{2*1}=\frac{11-3\sqrt{5}}{2} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+3\sqrt{5}}{2*1}=\frac{11+3\sqrt{5}}{2} $
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