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2m^2=15
We move all terms to the left:
2m^2-(15)=0
a = 2; b = 0; c = -15;
Δ = b2-4ac
Δ = 02-4·2·(-15)
Δ = 120
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}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{30}}{2*2}=\frac{0-2\sqrt{30}}{4} =-\frac{2\sqrt{30}}{4} =-\frac{\sqrt{30}}{2} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{30}}{2*2}=\frac{0+2\sqrt{30}}{4} =\frac{2\sqrt{30}}{4} =\frac{\sqrt{30}}{2} $
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