2x^2+5(x-1)=x(x+1)

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

2x^2+5(x-1)=x(x+1)

We move all terms to the left:

2x^2+5(x-1)-(x(x+1))=0

We multiply parentheses

2x^2+5x-(x(x+1))-5=0


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

x(x+1)

We multiply parentheses

x^2+x

Back to the equation:

-(x^2+x)


We get rid of parentheses

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

We add all the numbers together, and all the variables

x^2+4x-5=0

a = 1; b = 4; c = -5;
Δ = b2-4ac
Δ = 42-4·1·(-5)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-6}{2*1}=\frac{-10}{2}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+6}{2*1}=\frac{2}{2}
=1
$