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(4)=F^2-6
We move all terms to the left:
(4)-(F^2-6)=0
We get rid of parentheses
-F^2+6+4=0
We add all the numbers together, and all the variables
-1F^2+10=0
a = -1; b = 0; c = +10;
Δ = b2-4ac
Δ = 02-4·(-1)·10
Δ = 40
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}$$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{10}}{2*-1}=\frac{0-2\sqrt{10}}{-2} =-\frac{2\sqrt{10}}{-2} =-\frac{\sqrt{10}}{-1} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{10}}{2*-1}=\frac{0+2\sqrt{10}}{-2} =\frac{2\sqrt{10}}{-2} =\frac{\sqrt{10}}{-1} $
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