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

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

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

We move all terms to the left:

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

We multiply parentheses

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


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

-7(x-4)-x

We add all the numbers together, and all the variables

-1x-7(x-4)

We multiply parentheses

-1x-7x+28

We add all the numbers together, and all the variables

-8x+28

Back to the equation:

-(-8x+28)


We get rid of parentheses

-8x^2+24x+8x-28+4=0

We add all the numbers together, and all the variables

-8x^2+32x-24=0

a = -8; b = 32; c = -24;
Δ = b2-4ac
Δ = 322-4·(-8)·(-24)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-16}{2*-8}=\frac{-48}{-16}
=+3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+16}{2*-8}=\frac{-16}{-16}
=1
$