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(-2)=3F^2+5F-1
We move all terms to the left:
(-2)-(3F^2+5F-1)=0
We add all the numbers together, and all the variables
-(3F^2+5F-1)-2=0
We get rid of parentheses
-3F^2-5F+1-2=0
We add all the numbers together, and all the variables
-3F^2-5F-1=0
a = -3; b = -5; c = -1;
Δ = b2-4ac
Δ = -52-4·(-3)·(-1)
Δ = 13
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}$$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{13}}{2*-3}=\frac{5-\sqrt{13}}{-6} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{13}}{2*-3}=\frac{5+\sqrt{13}}{-6} $
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