25-(3x+5)=2x(x+8)+x

Solution for 25-(3x+5)=2x(x+8)+x equation:

25-(3x+5)=2x(x+8)+x

We move all terms to the left:

25-(3x+5)-(2x(x+8)+x)=0

We get rid of parentheses

-3x-(2x(x+8)+x)-5+25=0


We calculate terms in parentheses: -(2x(x+8)+x), so:

2x(x+8)+x

We add all the numbers together, and all the variables

x+2x(x+8)

We multiply parentheses

2x^2+x+16x

We add all the numbers together, and all the variables

2x^2+17x

Back to the equation:

-(2x^2+17x)


We add all the numbers together, and all the variables

-3x-(2x^2+17x)+20=0

We get rid of parentheses

-2x^2-3x-17x+20=0

We add all the numbers together, and all the variables

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

a = -2; b = -20; c = +20;
Δ = b2-4ac
Δ = -202-4·(-2)·20
Δ = 560
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:

$\sqrt{\Delta}=\sqrt{560}=\sqrt{16*35}=\sqrt{16}*\sqrt{35}=4\sqrt{35}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{35}}{2*-2}=\frac{20-4\sqrt{35}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{35}}{2*-2}=\frac{20+4\sqrt{35}}{-4}
$