-8x(x-3)+4=-7x(x-4)-x

Solution for -8x(x-3)+4=-7x(x-4)-x equation:

-8x(x-3)+4=-7x(x-4)-x

We move all terms to the left:

-8x(x-3)+4-(-7x(x-4)-x)=0

We multiply parentheses

-8x^2+24x-(-7x(x-4)-x)+4=0


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

-7x(x-4)-x

We add all the numbers together, and all the variables

-1x-7x(x-4)

We multiply parentheses

-7x^2-1x+28x

We add all the numbers together, and all the variables

-7x^2+27x

Back to the equation:

-(-7x^2+27x)


We get rid of parentheses

-8x^2+7x^2-27x+24x+4=0

We add all the numbers together, and all the variables

-1x^2-3x+4=0

a = -1; b = -3; c = +4;
Δ = b2-4ac
Δ = -32-4·(-1)·4
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-5}{2*-1}=\frac{-2}{-2}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+5}{2*-1}=\frac{8}{-2}
=-4
$