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

Solution for (x^2-9)=4(x-3) equation:

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

We move all terms to the left:

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

We get rid of parentheses

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


We calculate terms in parentheses: -(4(x-3)), so:

4(x-3)

We multiply parentheses

4x-12

Back to the equation:

-(4x-12)


We get rid of parentheses

x^2-4x+12-9=0

We add all the numbers together, and all the variables

x^2-4x+3=0

a = 1; b = -4; c = +3;
Δ = b2-4ac
Δ = -42-4·1·3
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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