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(F)=2F^2-45
We move all terms to the left:
(F)-(2F^2-45)=0
We get rid of parentheses
-2F^2+F+45=0
a = -2; b = 1; c = +45;
Δ = b2-4ac
Δ = 12-4·(-2)·45
Δ = 361
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{361}=19$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-19}{2*-2}=\frac{-20}{-4} =+5 $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+19}{2*-2}=\frac{18}{-4} =-4+1/2 $
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