8r^2+48r+64=0

Solution for 8r^2+48r+64=0 equation:

8r^2+48r+64=0

a = 8; b = 48; c = +64;
Δ = b2-4ac
Δ = 482-4·8·64
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-16}{2*8}=\frac{-64}{16}
=-4
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+16}{2*8}=\frac{-32}{16}
=-2
$