3b^2-8b-24b+63=0

Solution for 3b^2-8b-24b+63=0 equation:

3b^2-8b-24b+63=0

We add all the numbers together, and all the variables

3b^2-32b+63=0

a = 3; b = -32; c = +63;
Δ = b2-4ac
Δ = -322-4·3·63
Δ = 268
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{268}=\sqrt{4*67}=\sqrt{4}*\sqrt{67}=2\sqrt{67}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-2\sqrt{67}}{2*3}=\frac{32-2\sqrt{67}}{6}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+2\sqrt{67}}{2*3}=\frac{32+2\sqrt{67}}{6}
$