8x(3-2x)+5x(3x+1)=8x(1-x)

Solution for 8x(3-2x)+5x(3x+1)=8x(1-x) equation:

8x(3-2x)+5x(3x+1)=8x(1-x)

We move all terms to the left:

8x(3-2x)+5x(3x+1)-(8x(1-x))=0

We add all the numbers together, and all the variables

8x(-2x+3)+5x(3x+1)-(8x(-1x+1))=0

We multiply parentheses

-16x^2+15x^2+24x+5x-(8x(-1x+1))=0


We calculate terms in parentheses: -(8x(-1x+1)), so:

8x(-1x+1)

We multiply parentheses

-8x^2+8x

Back to the equation:

-(-8x^2+8x)


We add all the numbers together, and all the variables

-1x^2-(-8x^2+8x)+29x=0

We get rid of parentheses

-1x^2+8x^2-8x+29x=0

We add all the numbers together, and all the variables

7x^2+21x=0

a = 7; b = 21; c = 0;
Δ = b2-4ac
Δ = 212-4·7·0
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-21}{2*7}=\frac{-42}{14}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+21}{2*7}=\frac{0}{14}
=0
$