8x^2+48x-40x^2=-4

Solution for 8x^2+48x-40x^2=-4 equation:

8x^2+48x-40x^2=-4

We move all terms to the left:

8x^2+48x-40x^2-(-4)=0

We add all the numbers together, and all the variables

-32x^2+48x+4=0

a = -32; b = 48; c = +4;
Δ = b2-4ac
Δ = 482-4·(-32)·4
Δ = 2816
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:

$\sqrt{\Delta}=\sqrt{2816}=\sqrt{256*11}=\sqrt{256}*\sqrt{11}=16\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-16\sqrt{11}}{2*-32}=\frac{-48-16\sqrt{11}}{-64}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+16\sqrt{11}}{2*-32}=\frac{-48+16\sqrt{11}}{-64}
$