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a^2+22a+40=0
a = 1; b = 22; c = +40;
Δ = b2-4ac
Δ = 222-4·1·40
Δ = 324
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{324}=18$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-18}{2*1}=\frac{-40}{2} =-20 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+18}{2*1}=\frac{-4}{2} =-2 $
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