x(x+3)+3(x2+x)=20

Solution for x(x+3)+3(x2+x)=20 equation:

x(x+3)+3(x2+x)=20

We move all terms to the left:

x(x+3)+3(x2+x)-(20)=0

We add all the numbers together, and all the variables

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

We multiply parentheses

3x^2+x^2+3x+3x-20=0

We add all the numbers together, and all the variables

4x^2+6x-20=0

a = 4; b = 6; c = -20;
Δ = b2-4ac
Δ = 62-4·4·(-20)
Δ = 356
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:

$\sqrt{\Delta}=\sqrt{356}=\sqrt{4*89}=\sqrt{4}*\sqrt{89}=2\sqrt{89}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{89}}{2*4}=\frac{-6-2\sqrt{89}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{89}}{2*4}=\frac{-6+2\sqrt{89}}{8}
$