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X2+5X-486=0
We add all the numbers together, and all the variables
X^2+5X-486=0
a = 1; b = 5; c = -486;
Δ = b2-4ac
Δ = 52-4·1·(-486)
Δ = 1969
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{1969}}{2*1}=\frac{-5-\sqrt{1969}}{2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{1969}}{2*1}=\frac{-5+\sqrt{1969}}{2} $
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