65=(8x-2)(2x+7)

Solution for 65=(8x-2)(2x+7) equation:

65=(8x-2)(2x+7)

We move all terms to the left:

65-((8x-2)(2x+7))=0

We multiply parentheses ..

-((+16x^2+56x-4x-14))+65=0


We calculate terms in parentheses: -((+16x^2+56x-4x-14)), so:

(+16x^2+56x-4x-14)

We get rid of parentheses

16x^2+56x-4x-14

We add all the numbers together, and all the variables

16x^2+52x-14

Back to the equation:

-(16x^2+52x-14)


We get rid of parentheses

-16x^2-52x+14+65=0

We add all the numbers together, and all the variables

-16x^2-52x+79=0

a = -16; b = -52; c = +79;
Δ = b2-4ac
Δ = -522-4·(-16)·79
Δ = 7760
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:

$\sqrt{\Delta}=\sqrt{7760}=\sqrt{16*485}=\sqrt{16}*\sqrt{485}=4\sqrt{485}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-52)-4\sqrt{485}}{2*-16}=\frac{52-4\sqrt{485}}{-32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-52)+4\sqrt{485}}{2*-16}=\frac{52+4\sqrt{485}}{-32}
$