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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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


We calculate terms in parentheses: -((-1x+4)2), so:

(-1x+4)2

We multiply parentheses

-2x+8

Back to the equation:

-(-2x+8)


We get rid of parentheses

x^2-2x+3x+2x-6-8=0

We add all the numbers together, and all the variables

x^2+3x-14=0

a = 1; b = 3; c = -14;
Δ = b2-4ac
Δ = 32-4·1·(-14)
Δ = 65
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{65}}{2*1}=\frac{-3-\sqrt{65}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{65}}{2*1}=\frac{-3+\sqrt{65}}{2}
$