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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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


We calculate terms in parentheses: -(4(x^2-3)), so:

4(x^2-3)

We multiply parentheses

4x^2-12

Back to the equation:

-(4x^2-12)


We get rid of parentheses

x^2-4x^2-2x-3x+6+12=0

We add all the numbers together, and all the variables

-3x^2-5x+18=0

a = -3; b = -5; c = +18;
Δ = b2-4ac
Δ = -52-4·(-3)·18
Δ = 241
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{-(-5)-\sqrt{241}}{2*-3}=\frac{5-\sqrt{241}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{241}}{2*-3}=\frac{5+\sqrt{241}}{-6}
$