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

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

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

We move all terms to the left:

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


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

3(x+2)

We multiply parentheses

3x+6

Back to the equation:

-(3x+6)


We get rid of parentheses

x^2-3x-6-4=0

We add all the numbers together, and all the variables

x^2-3x-10=0

a = 1; b = -3; c = -10;
Δ = b2-4ac
Δ = -32-4·1·(-10)
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{-(-3)-7}{2*1}=\frac{-4}{2}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+7}{2*1}=\frac{10}{2}
=5
$