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

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

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

We move all terms to the left:

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

We multiply parentheses

-(2x^2-5(5-x^2-x))+2x^2-2x-10+5=0


We calculate terms in parentheses: -(2x^2-5(5-x^2-x)), so:

2x^2-5(5-x^2-x)

We multiply parentheses

2x^2+5x^2+5x-25

We add all the numbers together, and all the variables

7x^2+5x-25

Back to the equation:

-(7x^2+5x-25)


We add all the numbers together, and all the variables

2x^2-2x-(7x^2+5x-25)-5=0

We get rid of parentheses

2x^2-7x^2-2x-5x+25-5=0

We add all the numbers together, and all the variables

-5x^2-7x+20=0

a = -5; b = -7; c = +20;
Δ = b2-4ac
Δ = -72-4·(-5)·20
Δ = 449
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{-(-7)-\sqrt{449}}{2*-5}=\frac{7-\sqrt{449}}{-10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+\sqrt{449}}{2*-5}=\frac{7+\sqrt{449}}{-10}
$