x+2=(3x^2)+17x+20

Solution for x+2=(3x^2)+17x+20 equation:

x+2=(3x^2)+17x+20

We move all terms to the left:

x+2-((3x^2)+17x+20)=0

We get rid of parentheses

-3x^2+x-17x-20+2=0

We add all the numbers together, and all the variables

-3x^2-16x-18=0

a = -3; b = -16; c = -18;
Δ = b2-4ac
Δ = -162-4·(-3)·(-18)
Δ = 40
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:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2\sqrt{10}}{2*-3}=\frac{16-2\sqrt{10}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2\sqrt{10}}{2*-3}=\frac{16+2\sqrt{10}}{-6}
$