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a^2-13a+16=0
a = 1; b = -13; c = +16;
Δ = b2-4ac
Δ = -132-4·1·16
Δ = 105
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-\sqrt{105}}{2*1}=\frac{13-\sqrt{105}}{2} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{105}}{2*1}=\frac{13+\sqrt{105}}{2} $
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