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8m^2=14
We move all terms to the left:
8m^2-(14)=0
a = 8; b = 0; c = -14;
Δ = b2-4ac
Δ = 02-4·8·(-14)
Δ = 448
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}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{448}=\sqrt{64*7}=\sqrt{64}*\sqrt{7}=8\sqrt{7}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{7}}{2*8}=\frac{0-8\sqrt{7}}{16} =-\frac{8\sqrt{7}}{16} =-\frac{\sqrt{7}}{2} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{7}}{2*8}=\frac{0+8\sqrt{7}}{16} =\frac{8\sqrt{7}}{16} =\frac{\sqrt{7}}{2} $
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