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m^2=51
We move all terms to the left:
m^2-(51)=0
a = 1; b = 0; c = -51;
Δ = b2-4ac
Δ = 02-4·1·(-51)
Δ = 204
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}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{204}=\sqrt{4*51}=\sqrt{4}*\sqrt{51}=2\sqrt{51}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{51}}{2*1}=\frac{0-2\sqrt{51}}{2} =-\frac{2\sqrt{51}}{2} =-\sqrt{51} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{51}}{2*1}=\frac{0+2\sqrt{51}}{2} =\frac{2\sqrt{51}}{2} =\sqrt{51} $
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