2x-3x^2=x(5+x)

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

2x-3x^2=x(5+x)

We move all terms to the left:

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

We add all the numbers together, and all the variables

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


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

x(x+5)

We multiply parentheses

x^2+5x

Back to the equation:

-(x^2+5x)


We get rid of parentheses

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

We add all the numbers together, and all the variables

-4x^2-3x=0

a = -4; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·(-4)·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*-4}=\frac{0}{-8}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*-4}=\frac{6}{-8}
=-3/4
$