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(x)^2-13x+28=0
a = 1; b = -13; c = +28;
Δ = b2-4ac
Δ = -132-4·1·28
Δ = 57
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{-(-13)-\sqrt{57}}{2*1}=\frac{13-\sqrt{57}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{57}}{2*1}=\frac{13+\sqrt{57}}{2} $
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