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(F)=3F^2-15
We move all terms to the left:
(F)-(3F^2-15)=0
We get rid of parentheses
-3F^2+F+15=0
a = -3; b = 1; c = +15;
Δ = b2-4ac
Δ = 12-4·(-3)·15
Δ = 181
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{181}}{2*-3}=\frac{-1-\sqrt{181}}{-6} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{181}}{2*-3}=\frac{-1+\sqrt{181}}{-6} $
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