(4/5)(3x-7)=16

Solution for (4/5)(3x-7)=16 equation:

(4/5)(3x-7)=16

We move all terms to the left:

(4/5)(3x-7)-(16)=0


Domain of the equation: 5)(3x-7)!=0

x∈R


We add all the numbers together, and all the variables

(+4/5)(3x-7)-16=0

We multiply parentheses ..

(+12x^2+4/5*-7)-16=0

We multiply all the terms by the denominator


(+12x^2+4-16*5*-7)=0

We get rid of parentheses

12x^2+4-7-16*5*=0

We add all the numbers together, and all the variables

12x^2=0

a = 12; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·12·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{0}{24}=0$