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(F)=-2F^2+32
We move all terms to the left:
(F)-(-2F^2+32)=0
We get rid of parentheses
2F^2+F-32=0
a = 2; b = 1; c = -32;
Δ = b2-4ac
Δ = 12-4·2·(-32)
Δ = 257
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{257}}{2*2}=\frac{-1-\sqrt{257}}{4} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{257}}{2*2}=\frac{-1+\sqrt{257}}{4} $
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