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(F)=-F^2-5F-5
We move all terms to the left:
(F)-(-F^2-5F-5)=0
We get rid of parentheses
F^2+5F+F+5=0
We add all the numbers together, and all the variables
F^2+6F+5=0
a = 1; b = 6; c = +5;
Δ = b2-4ac
Δ = 62-4·1·5
Δ = 16
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{16}=4$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-4}{2*1}=\frac{-10}{2} =-5 $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+4}{2*1}=\frac{-2}{2} =-1 $
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