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m2-3m-10=03
We move all terms to the left:
m2-3m-10-(03)=0
We add all the numbers together, and all the variables
m^2-3m-13=0
a = 1; b = -3; c = -13;
Δ = b2-4ac
Δ = -32-4·1·(-13)
Δ = 61
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}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{61}}{2*1}=\frac{3-\sqrt{61}}{2} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{61}}{2*1}=\frac{3+\sqrt{61}}{2} $
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