(X^2)+6x+20=31-4x

Solution for (X^2)+6x+20=31-4x equation:

(X^2)+6X+20=31-4X

We move all terms to the left:

(X^2)+6X+20-(31-4X)=0

We add all the numbers together, and all the variables

X^2+6X-(-4X+31)+20=0

We get rid of parentheses

X^2+6X+4X-31+20=0

We add all the numbers together, and all the variables

X^2+10X-11=0

a = 1; b = 10; c = -11;
Δ = b2-4ac
Δ = 102-4·1·(-11)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-12}{2*1}=\frac{-22}{2}
=-11
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+12}{2*1}=\frac{2}{2}
=1
$