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m2-4m-21=0
We add all the numbers together, and all the variables
m^2-4m-21=0
a = 1; b = -4; c = -21;
Δ = b2-4ac
Δ = -42-4·1·(-21)
Δ = 100
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{100}=10$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-10}{2*1}=\frac{-6}{2} =-3 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+10}{2*1}=\frac{14}{2} =7 $
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