6^3-5b^2-4b+20=0

Solution for 6^3-5b^2-4b+20=0 equation:

6^3-5b^2-4b+20=0

We add all the numbers together, and all the variables

-5b^2-4b+236=0

a = -5; b = -4; c = +236;
Δ = b2-4ac
Δ = -42-4·(-5)·236
Δ = 4736
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:

$\sqrt{\Delta}=\sqrt{4736}=\sqrt{64*74}=\sqrt{64}*\sqrt{74}=8\sqrt{74}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8\sqrt{74}}{2*-5}=\frac{4-8\sqrt{74}}{-10}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8\sqrt{74}}{2*-5}=\frac{4+8\sqrt{74}}{-10}
$