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X^2+13X-19=0
a = 1; b = 13; c = -19;
Δ = b2-4ac
Δ = 132-4·1·(-19)
Δ = 245
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}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{245}=\sqrt{49*5}=\sqrt{49}*\sqrt{5}=7\sqrt{5}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-7\sqrt{5}}{2*1}=\frac{-13-7\sqrt{5}}{2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+7\sqrt{5}}{2*1}=\frac{-13+7\sqrt{5}}{2} $
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