8u^2+32u+24=0

Solution for 8u^2+32u+24=0 equation:

8u^2+32u+24=0

a = 8; b = 32; c = +24;
Δ = b2-4ac
Δ = 322-4·8·24
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-16}{2*8}=\frac{-48}{16}
=-3
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+16}{2*8}=\frac{-16}{16}
=-1
$