4x(x-3)+6=20

Solution for 4x(x-3)+6=20 equation:

4x(x-3)+6=20

We move all terms to the left:

4x(x-3)+6-(20)=0

We add all the numbers together, and all the variables

4x(x-3)-14=0

We multiply parentheses

4x^2-12x-14=0

a = 4; b = -12; c = -14;
Δ = b2-4ac
Δ = -122-4·4·(-14)
Δ = 368
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:

$\sqrt{\Delta}=\sqrt{368}=\sqrt{16*23}=\sqrt{16}*\sqrt{23}=4\sqrt{23}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{23}}{2*4}=\frac{12-4\sqrt{23}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{23}}{2*4}=\frac{12+4\sqrt{23}}{8}
$