3(x-3)(x-7)=4(x+3)

Solution for 3(x-3)(x-7)=4(x+3) equation:

3(x-3)(x-7)=4(x+3)

We move all terms to the left:

3(x-3)(x-7)-(4(x+3))=0

We multiply parentheses ..

3(+x^2-7x-3x+21)-(4(x+3))=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 multiply parentheses

3x^2-21x-9x-(4x+12)+63=0

We get rid of parentheses

3x^2-21x-9x-4x-12+63=0

We add all the numbers together, and all the variables

3x^2-34x+51=0

a = 3; b = -34; c = +51;
Δ = b2-4ac
Δ = -342-4·3·51
Δ = 544
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{544}=\sqrt{16*34}=\sqrt{16}*\sqrt{34}=4\sqrt{34}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-34)-4\sqrt{34}}{2*3}=\frac{34-4\sqrt{34}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-34)+4\sqrt{34}}{2*3}=\frac{34+4\sqrt{34}}{6}
$