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

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

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

We move all terms to the left:

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

We get rid of parentheses

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


We calculate terms in parentheses: -((3x^2+4x)-(x^2+7x)), so:

(3x^2+4x)-(x^2+7x)

We get rid of parentheses

3x^2-x^2+4x-7x

We add all the numbers together, and all the variables

2x^2-3x

Back to the equation:

-(2x^2-3x)


We add all the numbers together, and all the variables

-1x-(2x^2-3x)+64=0

We get rid of parentheses

-2x^2-1x+3x+64=0

We add all the numbers together, and all the variables

-2x^2+2x+64=0

a = -2; b = 2; c = +64;
Δ = b2-4ac
Δ = 22-4·(-2)·64
Δ = 516
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:

$\sqrt{\Delta}=\sqrt{516}=\sqrt{4*129}=\sqrt{4}*\sqrt{129}=2\sqrt{129}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{129}}{2*-2}=\frac{-2-2\sqrt{129}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{129}}{2*-2}=\frac{-2+2\sqrt{129}}{-4}
$