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m^2-29m+204=0
a = 1; b = -29; c = +204;
Δ = b2-4ac
Δ = -292-4·1·204
Δ = 25
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{25}=5$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-5}{2*1}=\frac{24}{2} =12 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+5}{2*1}=\frac{34}{2} =17 $
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