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(x)^2-13x-28=0
a = 1; b = -13; c = -28;
Δ = b2-4ac
Δ = -132-4·1·(-28)
Δ = 281
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{281}}{2*1}=\frac{13-\sqrt{281}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{281}}{2*1}=\frac{13+\sqrt{281}}{2} $
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