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2m^2-48=0
a = 2; b = 0; c = -48;
Δ = b2-4ac
Δ = 02-4·2·(-48)
Δ = 384
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}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{384}=\sqrt{64*6}=\sqrt{64}*\sqrt{6}=8\sqrt{6}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{6}}{2*2}=\frac{0-8\sqrt{6}}{4} =-\frac{8\sqrt{6}}{4} =-2\sqrt{6} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{6}}{2*2}=\frac{0+8\sqrt{6}}{4} =\frac{8\sqrt{6}}{4} =2\sqrt{6} $
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