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8+2m^2=158
We move all terms to the left:
8+2m^2-(158)=0
We add all the numbers together, and all the variables
2m^2-150=0
a = 2; b = 0; c = -150;
Δ = b2-4ac
Δ = 02-4·2·(-150)
Δ = 1200
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}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1200}=\sqrt{400*3}=\sqrt{400}*\sqrt{3}=20\sqrt{3}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{3}}{2*2}=\frac{0-20\sqrt{3}}{4} =-\frac{20\sqrt{3}}{4} =-5\sqrt{3} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{3}}{2*2}=\frac{0+20\sqrt{3}}{4} =\frac{20\sqrt{3}}{4} =5\sqrt{3} $
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