(x+3)2+(4-x2)=2(x-4)(x+3)

Solution for (x+3)2+(4-x2)=2(x-4)(x+3) equation:

(x+3)2+(4-x2)=2(x-4)(x+3)

We move all terms to the left:

(x+3)2+(4-x2)-(2(x-4)(x+3))=0

We add all the numbers together, and all the variables

(-1x^2+4)+(x+3)2-(2(x-4)(x+3))=0

We multiply parentheses

(-1x^2+4)+2x-(2(x-4)(x+3))+6=0

We get rid of parentheses

-1x^2+2x-(2(x-4)(x+3))+4+6=0

We multiply parentheses ..

-1x^2-(2(+x^2+3x-4x-12))+2x+4+6=0


We calculate terms in parentheses: -(2(+x^2+3x-4x-12)), so:

2(+x^2+3x-4x-12)

We multiply parentheses

2x^2+6x-8x-24

We add all the numbers together, and all the variables

2x^2-2x-24

Back to the equation:

-(2x^2-2x-24)


We add all the numbers together, and all the variables

-1x^2+2x-(2x^2-2x-24)+10=0

We get rid of parentheses

-1x^2-2x^2+2x+2x+24+10=0

We add all the numbers together, and all the variables

-3x^2+4x+34=0

a = -3; b = 4; c = +34;
Δ = b2-4ac
Δ = 42-4·(-3)·34
Δ = 424
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:

$\sqrt{\Delta}=\sqrt{424}=\sqrt{4*106}=\sqrt{4}*\sqrt{106}=2\sqrt{106}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{106}}{2*-3}=\frac{-4-2\sqrt{106}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{106}}{2*-3}=\frac{-4+2\sqrt{106}}{-6}
$