X2+2x=3(2-x-x2)

Solution for X2+2x=3(2-x-x2) equation:

X2+2X=3(2-X-X2)

We move all terms to the left:

X2+2X-(3(2-X-X2))=0

We add all the numbers together, and all the variables

-(3(-1X-1X^2+2))+X2+2X=0

We add all the numbers together, and all the variables

X^2-(3(-1X-1X^2+2))+2X=0


We calculate terms in parentheses: -(3(-1X-1X^2+2)), so:

3(-1X-1X^2+2)

We multiply parentheses

-3X^2-3X+6

Back to the equation:

-(-3X^2-3X+6)


We get rid of parentheses

X^2+3X^2+3X+2X-6=0

We add all the numbers together, and all the variables

4X^2+5X-6=0

a = 4; b = 5; c = -6;
Δ = b2-4ac
Δ = 52-4·4·(-6)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-11}{2*4}=\frac{-16}{8}
=-2
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+11}{2*4}=\frac{6}{8}
=3/4
$