4x^2-32x-1536=0

Solution for 4x^2-32x-1536=0 equation:

4x^2-32x-1536=0

a = 4; b = -32; c = -1536;
Δ = b2-4ac
Δ = -322-4·4·(-1536)
Δ = 25600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25600}=160$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-160}{2*4}=\frac{-128}{8}
=-16
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+160}{2*4}=\frac{192}{8}
=24
$