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(F)=49F^2-36
We move all terms to the left:
(F)-(49F^2-36)=0
We get rid of parentheses
-49F^2+F+36=0
a = -49; b = 1; c = +36;
Δ = b2-4ac
Δ = 12-4·(-49)·36
Δ = 7057
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{-(1)-\sqrt{7057}}{2*-49}=\frac{-1-\sqrt{7057}}{-98} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{7057}}{2*-49}=\frac{-1+\sqrt{7057}}{-98} $
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