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

Solution for (x-3)(x+4)=(x+2)2-1 equation:

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

We move all terms to the left:

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

We multiply parentheses ..

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


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

(x+2)2-1

We multiply parentheses

2x+4-1

We add all the numbers together, and all the variables

2x+3

Back to the equation:

-(2x+3)


We get rid of parentheses

x^2+4x-3x-2x-12-3=0

We add all the numbers together, and all the variables

x^2-1x-15=0

a = 1; b = -1; c = -15;
Δ = b2-4ac
Δ = -12-4·1·(-15)
Δ = 61
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{-(-1)-\sqrt{61}}{2*1}=\frac{1-\sqrt{61}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{61}}{2*1}=\frac{1+\sqrt{61}}{2}
$