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1-27m^2=0
a = -27; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-27)·1
Δ = 108
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}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{108}=\sqrt{36*3}=\sqrt{36}*\sqrt{3}=6\sqrt{3}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{3}}{2*-27}=\frac{0-6\sqrt{3}}{-54} =-\frac{6\sqrt{3}}{-54} =-\frac{\sqrt{3}}{-9} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{3}}{2*-27}=\frac{0+6\sqrt{3}}{-54} =\frac{6\sqrt{3}}{-54} =\frac{\sqrt{3}}{-9} $
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