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1+M2=134
We move all terms to the left:
1+M2-(134)=0
We add all the numbers together, and all the variables
M^2-133=0
a = 1; b = 0; c = -133;
Δ = b2-4ac
Δ = 02-4·1·(-133)
Δ = 532
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}$$M_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{532}=\sqrt{4*133}=\sqrt{4}*\sqrt{133}=2\sqrt{133}$$M_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{133}}{2*1}=\frac{0-2\sqrt{133}}{2} =-\frac{2\sqrt{133}}{2} =-\sqrt{133} $$M_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{133}}{2*1}=\frac{0+2\sqrt{133}}{2} =\frac{2\sqrt{133}}{2} =\sqrt{133} $
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