3x-x(4x+5)=-x(x+4)-9x-5

Solution for 3x-x(4x+5)=-x(x+4)-9x-5 equation:

3x-x(4x+5)=-x(x+4)-9x-5

We move all terms to the left:

3x-x(4x+5)-(-x(x+4)-9x-5)=0

We multiply parentheses

-4x^2+3x-5x-(-x(x+4)-9x-5)=0


We calculate terms in parentheses: -(-x(x+4)-9x-5), so:

-x(x+4)-9x-5

We add all the numbers together, and all the variables

-9x-x(x+4)-5

We multiply parentheses

-x^2-9x-4x-5

We add all the numbers together, and all the variables

-1x^2-13x-5

Back to the equation:

-(-1x^2-13x-5)


We add all the numbers together, and all the variables

-4x^2-(-1x^2-13x-5)-2x=0

We get rid of parentheses

-4x^2+1x^2+13x-2x+5=0

We add all the numbers together, and all the variables

-3x^2+11x+5=0

a = -3; b = 11; c = +5;
Δ = b2-4ac
Δ = 112-4·(-3)·5
Δ = 181
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{-(11)-\sqrt{181}}{2*-3}=\frac{-11-\sqrt{181}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{181}}{2*-3}=\frac{-11+\sqrt{181}}{-6}
$