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m^2-10m=5
We move all terms to the left:
m^2-10m-(5)=0
a = 1; b = -10; c = -5;
Δ = b2-4ac
Δ = -102-4·1·(-5)
Δ = 120
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}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{30}}{2*1}=\frac{10-2\sqrt{30}}{2} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{30}}{2*1}=\frac{10+2\sqrt{30}}{2} $
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