2x(1x+3)=20

Solution for 2x(1x+3)=20 equation:

2x(1x+3)=20

We move all terms to the left:

2x(1x+3)-(20)=0

We add all the numbers together, and all the variables

2x(x+3)-20=0

We multiply parentheses

2x^2+6x-20=0

a = 2; b = 6; c = -20;
Δ = b2-4ac
Δ = 62-4·2·(-20)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-14}{2*2}=\frac{-20}{4}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+14}{2*2}=\frac{8}{4}
=2
$