9-4x=3+x2

Solution for 9-4x=3+x2 equation:

9-4x=3+x2

We move all terms to the left:

9-4x-(3+x2)=0

We add all the numbers together, and all the variables

-(+x^2+3)-4x+9=0

We get rid of parentheses

-x^2-4x-3+9=0

We add all the numbers together, and all the variables

-1x^2-4x+6=0

a = -1; b = -4; c = +6;
Δ = b2-4ac
Δ = -42-4·(-1)·6
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{-(-4)-2\sqrt{10}}{2*-1}=\frac{4-2\sqrt{10}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{10}}{2*-1}=\frac{4+2\sqrt{10}}{-2}
$