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x^2+53x-196=0
a = 1; b = 53; c = -196;
Δ = b2-4ac
Δ = 532-4·1·(-196)
Δ = 3593
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{-(53)-\sqrt{3593}}{2*1}=\frac{-53-\sqrt{3593}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(53)+\sqrt{3593}}{2*1}=\frac{-53+\sqrt{3593}}{2} $
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