8x^2+12x^2-32x-48=0

Solution for 8x^2+12x^2-32x-48=0 equation:

8x^2+12x^2-32x-48=0

We add all the numbers together, and all the variables

20x^2-32x-48=0

a = 20; b = -32; c = -48;
Δ = b2-4ac
Δ = -322-4·20·(-48)
Δ = 4864
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:

$\sqrt{\Delta}=\sqrt{4864}=\sqrt{256*19}=\sqrt{256}*\sqrt{19}=16\sqrt{19}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-16\sqrt{19}}{2*20}=\frac{32-16\sqrt{19}}{40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+16\sqrt{19}}{2*20}=\frac{32+16\sqrt{19}}{40}
$