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

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

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

We move all terms to the left:

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

We multiply parentheses

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


We calculate terms in parentheses: -(4x+2(x-4)-3), so:

4x+2(x-4)-3

We multiply parentheses

4x+2x-8-3

We add all the numbers together, and all the variables

6x-11

Back to the equation:

-(6x-11)


We get rid of parentheses

3x^2-5x-6x+11-1=0

We add all the numbers together, and all the variables

3x^2-11x+10=0

a = 3; b = -11; c = +10;
Δ = b2-4ac
Δ = -112-4·3·10
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-1}{2*3}=\frac{10}{6}
=1+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+1}{2*3}=\frac{12}{6}
=2
$