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(9)=3F^2
We move all terms to the left:
(9)-(3F^2)=0
a = -3; b = 0; c = +9;
Δ = b2-4ac
Δ = 02-4·(-3)·9
Δ = 108
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{108}=\sqrt{36*3}=\sqrt{36}*\sqrt{3}=6\sqrt{3}$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{3}}{2*-3}=\frac{0-6\sqrt{3}}{-6} =-\frac{6\sqrt{3}}{-6} =-\frac{\sqrt{3}}{-1} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{3}}{2*-3}=\frac{0+6\sqrt{3}}{-6} =\frac{6\sqrt{3}}{-6} =\frac{\sqrt{3}}{-1} $
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