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1+M2=180
We move all terms to the left:
1+M2-(180)=0
We add all the numbers together, and all the variables
M^2-179=0
a = 1; b = 0; c = -179;
Δ = b2-4ac
Δ = 02-4·1·(-179)
Δ = 716
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}$$M_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{716}=\sqrt{4*179}=\sqrt{4}*\sqrt{179}=2\sqrt{179}$$M_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{179}}{2*1}=\frac{0-2\sqrt{179}}{2} =-\frac{2\sqrt{179}}{2} =-\sqrt{179} $$M_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{179}}{2*1}=\frac{0+2\sqrt{179}}{2} =\frac{2\sqrt{179}}{2} =\sqrt{179} $
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