2x(3x-5)-1=-7(x-1)+x-4

Solution for 2x(3x-5)-1=-7(x-1)+x-4 equation:

2x(3x-5)-1=-7(x-1)+x-4

We move all terms to the left:

2x(3x-5)-1-(-7(x-1)+x-4)=0

We multiply parentheses

6x^2-10x-(-7(x-1)+x-4)-1=0


We calculate terms in parentheses: -(-7(x-1)+x-4), so:

-7(x-1)+x-4

We add all the numbers together, and all the variables

x-7(x-1)-4

We multiply parentheses

x-7x+7-4

We add all the numbers together, and all the variables

-6x+3

Back to the equation:

-(-6x+3)


We get rid of parentheses

6x^2-10x+6x-3-1=0

We add all the numbers together, and all the variables

6x^2-4x-4=0

a = 6; b = -4; c = -4;
Δ = b2-4ac
Δ = -42-4·6·(-4)
Δ = 112
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:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{7}}{2*6}=\frac{4-4\sqrt{7}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{7}}{2*6}=\frac{4+4\sqrt{7}}{12}
$