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(6)=5F^2+3F+6
We move all terms to the left:
(6)-(5F^2+3F+6)=0
We get rid of parentheses
-5F^2-3F-6+6=0
We add all the numbers together, and all the variables
-5F^2-3F=0
a = -5; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·(-5)·0
Δ = 9
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{9}=3$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*-5}=\frac{0}{-10} =0 $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*-5}=\frac{6}{-10} =-3/5 $
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