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

-32x^2-2x+224x-(-5x+6)+19

We get rid of parentheses

-32x^2-2x+224x+5x-6+19

We add all the numbers together, and all the variables

-32x^2+227x+13

Back to the equation:

-(-32x^2+227x+13)


We add all the numbers together, and all the variables

-(-32x^2+227x+13)-13x-27=0

We get rid of parentheses

32x^2-227x-13x-13-27=0

We add all the numbers together, and all the variables

32x^2-240x-40=0

a = 32; b = -240; c = -40;
Δ = b2-4ac
Δ = -2402-4·32·(-40)
Δ = 62720
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:

$\sqrt{\Delta}=\sqrt{62720}=\sqrt{12544*5}=\sqrt{12544}*\sqrt{5}=112\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-240)-112\sqrt{5}}{2*32}=\frac{240-112\sqrt{5}}{64}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-240)+112\sqrt{5}}{2*32}=\frac{240+112\sqrt{5}}{64}
$