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(F)=2F^2+6F-5
We move all terms to the left:
(F)-(2F^2+6F-5)=0
We get rid of parentheses
-2F^2+F-6F+5=0
We add all the numbers together, and all the variables
-2F^2-5F+5=0
a = -2; b = -5; c = +5;
Δ = b2-4ac
Δ = -52-4·(-2)·5
Δ = 65
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{-(-5)-\sqrt{65}}{2*-2}=\frac{5-\sqrt{65}}{-4} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{65}}{2*-2}=\frac{5+\sqrt{65}}{-4} $
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