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(a^2)+23a+20=0
a = 1; b = 23; c = +20;
Δ = b2-4ac
Δ = 232-4·1·20
Δ = 449
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{-(23)-\sqrt{449}}{2*1}=\frac{-23-\sqrt{449}}{2} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+\sqrt{449}}{2*1}=\frac{-23+\sqrt{449}}{2} $
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