64=2x(4^x-3)

Solution for 64=2x(4^x-3) equation:

64=2x(4^x-3)

We move all terms to the left:

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


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

2x(4^x-3)

We multiply parentheses

8x^2-6x

Back to the equation:

-(8x^2-6x)


We get rid of parentheses

-8x^2+6x+64=0

a = -8; b = 6; c = +64;
Δ = b2-4ac
Δ = 62-4·(-8)·64
Δ = 2084
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:

$\sqrt{\Delta}=\sqrt{2084}=\sqrt{4*521}=\sqrt{4}*\sqrt{521}=2\sqrt{521}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{521}}{2*-8}=\frac{-6-2\sqrt{521}}{-16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{521}}{2*-8}=\frac{-6+2\sqrt{521}}{-16}
$