0=(6x^2+20x)-(x^2+8x+64)

Solution for 0=(6x^2+20x)-(x^2+8x+64) equation:

0=(6x^2+20x)-(x^2+8x+64)

We move all terms to the left:

0-((6x^2+20x)-(x^2+8x+64))=0

We add all the numbers together, and all the variables

-((6x^2+20x)-(x^2+8x+64))=0


We calculate terms in parentheses: -((6x^2+20x)-(x^2+8x+64)), so:

(6x^2+20x)-(x^2+8x+64)

We get rid of parentheses

6x^2-x^2+20x-8x-64

We add all the numbers together, and all the variables

5x^2+12x-64

Back to the equation:

-(5x^2+12x-64)


We get rid of parentheses

-5x^2-12x+64=0

a = -5; b = -12; c = +64;
Δ = b2-4ac
Δ = -122-4·(-5)·64
Δ = 1424
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:

$\sqrt{\Delta}=\sqrt{1424}=\sqrt{16*89}=\sqrt{16}*\sqrt{89}=4\sqrt{89}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{89}}{2*-5}=\frac{12-4\sqrt{89}}{-10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{89}}{2*-5}=\frac{12+4\sqrt{89}}{-10}
$