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(F+8)=F^2-4F+3
We move all terms to the left:
(F+8)-(F^2-4F+3)=0
We get rid of parentheses
-F^2+F+4F+8-3=0
We add all the numbers together, and all the variables
-1F^2+5F+5=0
a = -1; b = 5; c = +5;
Δ = b2-4ac
Δ = 52-4·(-1)·5
Δ = 45
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}$$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{45}=\sqrt{9*5}=\sqrt{9}*\sqrt{5}=3\sqrt{5}$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-3\sqrt{5}}{2*-1}=\frac{-5-3\sqrt{5}}{-2} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+3\sqrt{5}}{2*-1}=\frac{-5+3\sqrt{5}}{-2} $
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