4x(12x+x)=48

Solution for 4x(12x+x)=48 equation:

4x(12x+x)=48

We move all terms to the left:

4x(12x+x)-(48)=0

We add all the numbers together, and all the variables

4x(+13x)-48=0

We multiply parentheses

52x^2-48=0

a = 52; b = 0; c = -48;
Δ = b2-4ac
Δ = 02-4·52·(-48)
Δ = 9984
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:

$\sqrt{\Delta}=\sqrt{9984}=\sqrt{256*39}=\sqrt{256}*\sqrt{39}=16\sqrt{39}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{39}}{2*52}=\frac{0-16\sqrt{39}}{104}
=-\frac{16\sqrt{39}}{104}
=-\frac{2\sqrt{39}}{13}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{39}}{2*52}=\frac{0+16\sqrt{39}}{104}
=\frac{16\sqrt{39}}{104}
=\frac{2\sqrt{39}}{13}
$