3200=0+(6.4)(222)+(1/2)a(222)^2

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Solution for 3200=0+(6.4)(222)+(1/2)a(222)^2 equation: 



3200=0+(6.4)(222)+(1/2)a(222)^2

We move all terms to the left:

3200-(0+(6.4)(222)+(1/2)a(222)^2)=0



Domain of the equation: 2)a222^2)!=0

a!=0/1

a!=0

a∈R



We add all the numbers together, and all the variables

-(0+(6.4)222+(+1/2)a222^2)+3200=0

We multiply all the terms by the denominator


-(0+(6.4)222+(+1+3200*2)a222^2)=0



We calculate terms in parentheses: -(0+(6.4)222+(+1+3200*2)a222^2), so:

0+(6.4)222+(+1+3200*2)a222^2

determiningTheFunctionDomain
(+1+3200*2)a222^2+0+(6.4)222

We add all the numbers together, and all the variables

6401a222^2+0+(6.4)222

We add all the numbers together, and all the variables

6401a222^2+1420.8

Back to the equation:

-(6401a222^2+1420.8)



We get rid of parentheses

-6401a222^2-1420.8=0

We move all terms containing a to the left, all other terms to the right

-6401a222^2=1420.8

a=1420.8/1

a=1420.8

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