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(F)=4F^2-11F-3
We move all terms to the left:
(F)-(4F^2-11F-3)=0
We get rid of parentheses
-4F^2+F+11F+3=0
We add all the numbers together, and all the variables
-4F^2+12F+3=0
a = -4; b = 12; c = +3;
Δ = b2-4ac
Δ = 122-4·(-4)·3
Δ = 192
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}$$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8\sqrt{3}}{2*-4}=\frac{-12-8\sqrt{3}}{-8} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8\sqrt{3}}{2*-4}=\frac{-12+8\sqrt{3}}{-8} $
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