4x^2+48x+128=

Solution for 4x^2+48x+128= equation:

4x^2+48x+128=

We move all terms to the left:

4x^2+48x+128-()=0

We add all the numbers together, and all the variables

4x^2+48x=0

a = 4; b = 48; c = 0;
Δ = b2-4ac
Δ = 482-4·4·0
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-48}{2*4}=\frac{-96}{8}
=-12
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+48}{2*4}=\frac{0}{8}
=0
$