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(F-3)=F^2-5F-8
We move all terms to the left:
(F-3)-(F^2-5F-8)=0
We get rid of parentheses
-F^2+F+5F-3+8=0
We add all the numbers together, and all the variables
-1F^2+6F+5=0
a = -1; b = 6; c = +5;
Δ = b2-4ac
Δ = 62-4·(-1)·5
Δ = 56
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}$$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{56}=\sqrt{4*14}=\sqrt{4}*\sqrt{14}=2\sqrt{14}$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{14}}{2*-1}=\frac{-6-2\sqrt{14}}{-2} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{14}}{2*-1}=\frac{-6+2\sqrt{14}}{-2} $
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