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13=5m^2
We move all terms to the left:
13-(5m^2)=0
a = -5; b = 0; c = +13;
Δ = b2-4ac
Δ = 02-4·(-5)·13
Δ = 260
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}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{260}=\sqrt{4*65}=\sqrt{4}*\sqrt{65}=2\sqrt{65}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{65}}{2*-5}=\frac{0-2\sqrt{65}}{-10} =-\frac{2\sqrt{65}}{-10} =-\frac{\sqrt{65}}{-5} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{65}}{2*-5}=\frac{0+2\sqrt{65}}{-10} =\frac{2\sqrt{65}}{-10} =\frac{\sqrt{65}}{-5} $
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