8(x^2-3x-4)=0

Solution for 8(x^2-3x-4)=0 equation:

8(x^2-3x-4)=0

We multiply parentheses

8x^2-24x-32=0

a = 8; b = -24; c = -32;
Δ = b2-4ac
Δ = -242-4·8·(-32)
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-40}{2*8}=\frac{-16}{16}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+40}{2*8}=\frac{64}{16}
=4
$