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a-5+a2=0
We add all the numbers together, and all the variables
a^2+a-5=0
a = 1; b = 1; c = -5;
Δ = b2-4ac
Δ = 12-4·1·(-5)
Δ = 21
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{21}}{2*1}=\frac{-1-\sqrt{21}}{2} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{21}}{2*1}=\frac{-1+\sqrt{21}}{2} $
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