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m^2+22m+21=0
a = 1; b = 22; c = +21;
Δ = b2-4ac
Δ = 222-4·1·21
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{400}=20$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-20}{2*1}=\frac{-42}{2} =-21 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+20}{2*1}=\frac{-2}{2} =-1 $
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