5(x-3)+7(2-x)-4(2x+7)-3+2=11-8(x-7)+4x(6-5x)+8-2x

Solution for 5(x-3)+7(2-x)-4(2x+7)-3+2=11-8(x-7)+4x(6-5x)+8-2x equation:

5(x-3)+7(2-x)-4(2x+7)-3+2=11-8(x-7)+4x(6-5x)+8-2x

We move all terms to the left:

5(x-3)+7(2-x)-4(2x+7)-3+2-(11-8(x-7)+4x(6-5x)+8-2x)=0

We add all the numbers together, and all the variables

5(x-3)+7(-1x+2)-4(2x+7)-(11-8(x-7)+4x(-5x+6)+8-2x)-3+2=0

We add all the numbers together, and all the variables

5(x-3)+7(-1x+2)-4(2x+7)-(11-8(x-7)+4x(-5x+6)+8-2x)-1=0

We multiply parentheses

5x-7x-8x-(11-8(x-7)+4x(-5x+6)+8-2x)-15+14-28-1=0


We calculate terms in parentheses: -(11-8(x-7)+4x(-5x+6)+8-2x), so:

11-8(x-7)+4x(-5x+6)+8-2x

determiningTheFunctionDomain
-8(x-7)+4x(-5x+6)-2x+11+8

We add all the numbers together, and all the variables

-2x-8(x-7)+4x(-5x+6)+19

We multiply parentheses

-20x^2-2x-8x+24x+56+19

We add all the numbers together, and all the variables

-20x^2+14x+75

Back to the equation:

-(-20x^2+14x+75)


We add all the numbers together, and all the variables

-(-20x^2+14x+75)-10x-30=0

We get rid of parentheses

20x^2-14x-10x-75-30=0

We add all the numbers together, and all the variables

20x^2-24x-105=0

a = 20; b = -24; c = -105;
Δ = b2-4ac
Δ = -242-4·20·(-105)
Δ = 8976
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:

$\sqrt{\Delta}=\sqrt{8976}=\sqrt{16*561}=\sqrt{16}*\sqrt{561}=4\sqrt{561}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-4\sqrt{561}}{2*20}=\frac{24-4\sqrt{561}}{40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+4\sqrt{561}}{2*20}=\frac{24+4\sqrt{561}}{40}
$