4(4x^2-33x+48)=0

Solution for 4(4x^2-33x+48)=0 equation:

4(4x^2-33x+48)=0

We multiply parentheses

16x^2-132x+192=0

a = 16; b = -132; c = +192;
Δ = b2-4ac
Δ = -1322-4·16·192
Δ = 5136
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:

$\sqrt{\Delta}=\sqrt{5136}=\sqrt{16*321}=\sqrt{16}*\sqrt{321}=4\sqrt{321}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-132)-4\sqrt{321}}{2*16}=\frac{132-4\sqrt{321}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-132)+4\sqrt{321}}{2*16}=\frac{132+4\sqrt{321}}{32}
$