-48-32x+16x^2=0

Solution for -48-32x+16x^2=0 equation:

-48-32x+16x^2=0

a = 16; b = -32; c = -48;
Δ = b2-4ac
Δ = -322-4·16·(-48)
Δ = 4096
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4096}=64$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-64}{2*16}=\frac{-32}{32}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+64}{2*16}=\frac{96}{32}
=3
$