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(F)=3F^2+2F-5
We move all terms to the left:
(F)-(3F^2+2F-5)=0
We get rid of parentheses
-3F^2+F-2F+5=0
We add all the numbers together, and all the variables
-3F^2-1F+5=0
a = -3; b = -1; c = +5;
Δ = b2-4ac
Δ = -12-4·(-3)·5
Δ = 61
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{-(-1)-\sqrt{61}}{2*-3}=\frac{1-\sqrt{61}}{-6} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{61}}{2*-3}=\frac{1+\sqrt{61}}{-6} $
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