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a^2-19a-18=0
a = 1; b = -19; c = -18;
Δ = b2-4ac
Δ = -192-4·1·(-18)
Δ = 433
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{-(-19)-\sqrt{433}}{2*1}=\frac{19-\sqrt{433}}{2} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{433}}{2*1}=\frac{19+\sqrt{433}}{2} $
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