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n2-10n+13=0
We add all the numbers together, and all the variables
n^2-10n+13=0
a = 1; b = -10; c = +13;
Δ = b2-4ac
Δ = -102-4·1·13
Δ = 48
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-4\sqrt{3}}{2*1}=\frac{10-4\sqrt{3}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+4\sqrt{3}}{2*1}=\frac{10+4\sqrt{3}}{2} $
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