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(F)=4F^2+4F-3
We move all terms to the left:
(F)-(4F^2+4F-3)=0
We get rid of parentheses
-4F^2+F-4F+3=0
We add all the numbers together, and all the variables
-4F^2-3F+3=0
a = -4; b = -3; c = +3;
Δ = b2-4ac
Δ = -32-4·(-4)·3
Δ = 57
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{57}}{2*-4}=\frac{3-\sqrt{57}}{-8} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{57}}{2*-4}=\frac{3+\sqrt{57}}{-8} $
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