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

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

(X-3)(X+4)=-^2+4

We move all terms to the left:

(X-3)(X+4)-(-^2+4)=0

We add all the numbers together, and all the variables

(X-3)(X+4)=0

We multiply parentheses ..

(+X^2+4X-3X-12)=0

We get rid of parentheses

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

We add all the numbers together, and all the variables

X^2+X-12=0

a = 1; b = 1; c = -12;
Δ = b2-4ac
Δ = 12-4·1·(-12)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-7}{2*1}=\frac{-8}{2}
=-4
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+7}{2*1}=\frac{6}{2}
=3
$