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7e^2=41
We move all terms to the left:
7e^2-(41)=0
a = 7; b = 0; c = -41;
Δ = b2-4ac
Δ = 02-4·7·(-41)
Δ = 1148
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$e_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$e_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1148}=\sqrt{4*287}=\sqrt{4}*\sqrt{287}=2\sqrt{287}$$e_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{287}}{2*7}=\frac{0-2\sqrt{287}}{14} =-\frac{2\sqrt{287}}{14} =-\frac{\sqrt{287}}{7} $$e_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{287}}{2*7}=\frac{0+2\sqrt{287}}{14} =\frac{2\sqrt{287}}{14} =\frac{\sqrt{287}}{7} $
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