16r^2+24r+8=0

Solution for 16r^2+24r+8=0 equation:

16r^2+24r+8=0

a = 16; b = 24; c = +8;
Δ = b2-4ac
Δ = 242-4·16·8
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-8}{2*16}=\frac{-32}{32}
=-1
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+8}{2*16}=\frac{-16}{32}
=-1/2
$