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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+3F+5=0
a = -3; b = 3; c = +5;
Δ = b2-4ac
Δ = 32-4·(-3)·5
Δ = 69
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{-(3)-\sqrt{69}}{2*-3}=\frac{-3-\sqrt{69}}{-6} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{69}}{2*-3}=\frac{-3+\sqrt{69}}{-6} $
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