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(F)=4(F+5)(2F-1)
We move all terms to the left:
(F)-(4(F+5)(2F-1))=0
We multiply parentheses ..
-(4(+2F^2-1F+10F-5))+F=0
We calculate terms in parentheses: -(4(+2F^2-1F+10F-5)), so:We add all the numbers together, and all the variables
4(+2F^2-1F+10F-5)
We multiply parentheses
8F^2-4F+40F-20
We add all the numbers together, and all the variables
8F^2+36F-20
Back to the equation:
-(8F^2+36F-20)
F-(8F^2+36F-20)=0
We get rid of parentheses
-8F^2+F-36F+20=0
We add all the numbers together, and all the variables
-8F^2-35F+20=0
a = -8; b = -35; c = +20;
Δ = b2-4ac
Δ = -352-4·(-8)·20
Δ = 1865
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{-(-35)-\sqrt{1865}}{2*-8}=\frac{35-\sqrt{1865}}{-16} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+\sqrt{1865}}{2*-8}=\frac{35+\sqrt{1865}}{-16} $
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