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(0)=3F^2-6F-45
We move all terms to the left:
(0)-(3F^2-6F-45)=0
We add all the numbers together, and all the variables
-(3F^2-6F-45)=0
We get rid of parentheses
-3F^2+6F+45=0
a = -3; b = 6; c = +45;
Δ = b2-4ac
Δ = 62-4·(-3)·45
Δ = 576
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}$$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{576}=24$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-24}{2*-3}=\frac{-30}{-6} =+5 $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+24}{2*-3}=\frac{18}{-6} =-3 $
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