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(3)=5F^2-2F
We move all terms to the left:
(3)-(5F^2-2F)=0
We get rid of parentheses
-5F^2+2F+3=0
a = -5; b = 2; c = +3;
Δ = b2-4ac
Δ = 22-4·(-5)·3
Δ = 64
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{64}=8$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-8}{2*-5}=\frac{-10}{-10} =1 $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+8}{2*-5}=\frac{6}{-10} =-3/5 $
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