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a2+9a-22=0
We add all the numbers together, and all the variables
a^2+9a-22=0
a = 1; b = 9; c = -22;
Δ = b2-4ac
Δ = 92-4·1·(-22)
Δ = 169
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}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{169}=13$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-13}{2*1}=\frac{-22}{2} =-11 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+13}{2*1}=\frac{4}{2} =2 $
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