20(3+a)=(7a+1/6)20

Solution for 20(3+a)=(7a+1/6)20 equation:

20(3+a)=(7a+1/6)20

We move all terms to the left:

20(3+a)-((7a+1/6)20)=0

We add all the numbers together, and all the variables

20(a+3)-((+7a+1/6)20)=0

We multiply parentheses

20a-((+7a+1/6)20)+60=0

We multiply all the terms by the denominator


20a*6)20)-((+7a+1+60*6)20)=0

We add all the numbers together, and all the variables

20a*6)20)-((7a+361)20)=0

We add all the numbers together, and all the variables

20a*6)20)-((7a=0

Wy multiply elements

120a^2=0

a = 120; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·120·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$a=\frac{-b}{2a}=\frac{0}{240}=0$