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m^2-12m-64=0
a = 1; b = -12; c = -64;
Δ = b2-4ac
Δ = -122-4·1·(-64)
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$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-20}{2*1}=\frac{-8}{2} =-4 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+20}{2*1}=\frac{32}{2} =16 $
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