a^2+(70-10A)+(160-5a)=180

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Solution for a^2+(70-10A)+(160-5a)=180 equation: 



a^2+(70-10)+(160-5a)=180

We move all terms to the left:

a^2+(70-10)+(160-5a)-(180)=0

determiningTheFunctionDomain
a^2+(160-5a)-180+(70-10)=0

We add all the numbers together, and all the variables

a^2+(-5a+160)-180+60=0

We add all the numbers together, and all the variables

a^2+(-5a+160)-120=0

We get rid of parentheses

a^2-5a+160-120=0

We add all the numbers together, and all the variables

a^2-5a+40=0

a = 1; b = -5; c = +40;
Δ = b2-4ac
Δ = -52-4·1·40
Δ = -135
Delta is less than zero, so there is no solution for the equation

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