(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))=0

We get rid of parentheses

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


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

(3x^2+4x)

We get rid of parentheses

3x^2+4x

Back to the equation:

-(3x^2+4x)


We add all the numbers together, and all the variables

-1x-(3x^2+4x)+64=0

We get rid of parentheses

-3x^2-1x-4x+64=0

We add all the numbers together, and all the variables

-3x^2-5x+64=0

a = -3; b = -5; c = +64;
Δ = b2-4ac
Δ = -52-4·(-3)·64
Δ = 793
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{-(-5)-\sqrt{793}}{2*-3}=\frac{5-\sqrt{793}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{793}}{2*-3}=\frac{5+\sqrt{793}}{-6}
$