3x+(5x-3)(2x-4)=-x(-x-4)-1

Solution for 3x+(5x-3)(2x-4)=-x(-x-4)-1 equation:

3x+(5x-3)(2x-4)=-x(-x-4)-1

We move all terms to the left:

3x+(5x-3)(2x-4)-(-x(-x-4)-1)=0

We add all the numbers together, and all the variables

3x+(5x-3)(2x-4)-(-x(-1x-4)-1)=0

We multiply parentheses ..

(+10x^2-20x-6x+12)+3x-(-x(-1x-4)-1)=0


We calculate terms in parentheses: -(-x(-1x-4)-1), so:

-x(-1x-4)-1

We multiply parentheses

1x^2+4x-1

We add all the numbers together, and all the variables

x^2+4x-1

Back to the equation:

-(x^2+4x-1)


We get rid of parentheses

10x^2-x^2-20x-6x+3x-4x+12+1=0

We add all the numbers together, and all the variables

9x^2-27x+13=0

a = 9; b = -27; c = +13;
Δ = b2-4ac
Δ = -272-4·9·13
Δ = 261
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:

$\sqrt{\Delta}=\sqrt{261}=\sqrt{9*29}=\sqrt{9}*\sqrt{29}=3\sqrt{29}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-27)-3\sqrt{29}}{2*9}=\frac{27-3\sqrt{29}}{18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+3\sqrt{29}}{2*9}=\frac{27+3\sqrt{29}}{18}
$