7x^2-(4x-3x^2-2x)-x+1=x+8

Solution for 7x^2-(4x-3x^2-2x)-x+1=x+8 equation:

7x^2-(4x-3x^2-2x)-x+1=x+8

We move all terms to the left:

7x^2-(4x-3x^2-2x)-x+1-(x+8)=0

We add all the numbers together, and all the variables

7x^2-(4x-3x^2-2x)-1x-(x+8)+1=0

We get rid of parentheses

7x^2+3x^2-4x+2x-1x-x-8+1=0

We add all the numbers together, and all the variables

10x^2-4x-7=0

a = 10; b = -4; c = -7;
Δ = b2-4ac
Δ = -42-4·10·(-7)
Δ = 296
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:

$\sqrt{\Delta}=\sqrt{296}=\sqrt{4*74}=\sqrt{4}*\sqrt{74}=2\sqrt{74}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{74}}{2*10}=\frac{4-2\sqrt{74}}{20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{74}}{2*10}=\frac{4+2\sqrt{74}}{20}
$