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m2+8m-65=0
We add all the numbers together, and all the variables
m^2+8m-65=0
a = 1; b = 8; c = -65;
Δ = b2-4ac
Δ = 82-4·1·(-65)
Δ = 324
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{324}=18$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-18}{2*1}=\frac{-26}{2} =-13 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+18}{2*1}=\frac{10}{2} =5 $
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