8/5(4.5-6)=-1/5(2z+3)

Solution for 8/5(4.5-6)=-1/5(2z+3) equation:

8/5(4.5-6)=-1/5(2z+3)

We move all terms to the left:

8/5(4.5-6)-(-1/5(2z+3))=0


Domain of the equation: 5(2z+3))!=0

z∈R


We add all the numbers together, and all the variables

-(-1/5(2z+3))+8/5(-1.5)=0

We calculate fractions


()/(5(2z+3))*5()+(40z2/(5(2z+3))*5()=0


We calculate terms in parentheses: +()/(5(2z+3))*5(), so:

)/(5(2z+3))*5(

We multiply all the terms by the denominator


)

We add all the numbers together, and all the variables



Back to the equation:

+()



We calculate terms in parentheses: +(40z2/(5(2z+3))*5(), so:

40z2/(5(2z+3))*5(

We multiply all the terms by the denominator


40z2

We add all the numbers together, and all the variables

40z^2

Back to the equation:

+(40z^2)


We add all the numbers together, and all the variables

40z^2=0

a = 40; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·40·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$z=\frac{-b}{2a}=\frac{0}{80}=0$