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X^2+5X-408=0
a = 1; b = 5; c = -408;
Δ = b2-4ac
Δ = 52-4·1·(-408)
Δ = 1657
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{-(5)-\sqrt{1657}}{2*1}=\frac{-5-\sqrt{1657}}{2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{1657}}{2*1}=\frac{-5+\sqrt{1657}}{2} $
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