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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses

-3x+5x-7x-8x-(11-8(x-7)+4x(-5x+6)+8-2x)-15+14-28+2=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)-13x-27=0

We get rid of parentheses

20x^2-14x-13x-75-27=0

We add all the numbers together, and all the variables

20x^2-27x-102=0

a = 20; b = -27; c = -102;
Δ = b2-4ac
Δ = -272-4·20·(-102)
Δ = 8889
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-27)-\sqrt{8889}}{2*20}=\frac{27-\sqrt{8889}}{40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+\sqrt{8889}}{2*20}=\frac{27+\sqrt{8889}}{40}
$