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x^2+9x-196=0
a = 1; b = 9; c = -196;
Δ = b2-4ac
Δ = 92-4·1·(-196)
Δ = 865
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-\sqrt{865}}{2*1}=\frac{-9-\sqrt{865}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{865}}{2*1}=\frac{-9+\sqrt{865}}{2} $
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