F(x)=6(3-x)(3-x)+10

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Solution for F(x)=6(3-x)(3-x)+10 equation:



(F)=6(3-F)(3-F)+10
We move all terms to the left:
(F)-(6(3-F)(3-F)+10)=0
We add all the numbers together, and all the variables
F-(6(-1F+3)(-1F+3)+10)=0
We multiply parentheses ..
-(6(+F^2-3F-3F+9)+10)+F=0
We calculate terms in parentheses: -(6(+F^2-3F-3F+9)+10), so:
6(+F^2-3F-3F+9)+10
We multiply parentheses
6F^2-18F-18F+54+10
We add all the numbers together, and all the variables
6F^2-36F+64
Back to the equation:
-(6F^2-36F+64)
We add all the numbers together, and all the variables
F-(6F^2-36F+64)=0
We get rid of parentheses
-6F^2+F+36F-64=0
We add all the numbers together, and all the variables
-6F^2+37F-64=0
a = -6; b = 37; c = -64;
Δ = b2-4ac
Δ = 372-4·(-6)·(-64)
Δ = -167
Delta is less than zero, so there is no solution for the equation

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