(3x-4)(x-20)-5=0

Solution for (3x-4)(x-20)-5=0 equation:

(3x-4)(x-20)-5=0

We multiply parentheses ..

(+3x^2-60x-4x+80)-5=0

We get rid of parentheses

3x^2-60x-4x+80-5=0

We add all the numbers together, and all the variables

3x^2-64x+75=0

a = 3; b = -64; c = +75;
Δ = b2-4ac
Δ = -642-4·3·75
Δ = 3196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{3196}=\sqrt{4*799}=\sqrt{4}*\sqrt{799}=2\sqrt{799}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-64)-2\sqrt{799}}{2*3}=\frac{64-2\sqrt{799}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-64)+2\sqrt{799}}{2*3}=\frac{64+2\sqrt{799}}{6}
$