3x^2+8x+24x+64=0

Solution for 3x^2+8x+24x+64=0 equation:

3x^2+8x+24x+64=0

We add all the numbers together, and all the variables

3x^2+32x+64=0

a = 3; b = 32; c = +64;
Δ = b2-4ac
Δ = 322-4·3·64
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-16}{2*3}=\frac{-48}{6}
=-8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+16}{2*3}=\frac{-16}{6}
=-2+2/3
$