4(3x-7)-7=1/5x-3

Solution for 4(3x-7)-7=1/5x-3 equation:

4(3x-7)-7=1/5x-3

We move all terms to the left:

4(3x-7)-7-(1/5x-3)=0


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

x∈R


We multiply parentheses

12x-(1/5x-3)-28-7=0

We get rid of parentheses

12x-1/5x+3-28-7=0

We multiply all the terms by the denominator


12x*5x+3*5x-28*5x-7*5x-1=0

Wy multiply elements

60x^2+15x-140x-35x-1=0

We add all the numbers together, and all the variables

60x^2-160x-1=0

a = 60; b = -160; c = -1;
Δ = b2-4ac
Δ = -1602-4·60·(-1)
Δ = 25840
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{25840}=\sqrt{16*1615}=\sqrt{16}*\sqrt{1615}=4\sqrt{1615}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-160)-4\sqrt{1615}}{2*60}=\frac{160-4\sqrt{1615}}{120}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-160)+4\sqrt{1615}}{2*60}=\frac{160+4\sqrt{1615}}{120}
$