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(F)=-16F^2+30F+6
We move all terms to the left:
(F)-(-16F^2+30F+6)=0
We get rid of parentheses
16F^2-30F+F-6=0
We add all the numbers together, and all the variables
16F^2-29F-6=0
a = 16; b = -29; c = -6;
Δ = b2-4ac
Δ = -292-4·16·(-6)
Δ = 1225
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{1225}=35$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-35}{2*16}=\frac{-6}{32} =-3/16 $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+35}{2*16}=\frac{64}{32} =2 $
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