3x(x-4)=8(x+3)+4

Solution for 3x(x-4)=8(x+3)+4 equation:

3x(x-4)=8(x+3)+4

We move all terms to the left:

3x(x-4)-(8(x+3)+4)=0

We multiply parentheses

3x^2-12x-(8(x+3)+4)=0


We calculate terms in parentheses: -(8(x+3)+4), so:

8(x+3)+4

We multiply parentheses

8x+24+4

We add all the numbers together, and all the variables

8x+28

Back to the equation:

-(8x+28)


We get rid of parentheses

3x^2-12x-8x-28=0

We add all the numbers together, and all the variables

3x^2-20x-28=0

a = 3; b = -20; c = -28;
Δ = b2-4ac
Δ = -202-4·3·(-28)
Δ = 736
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:

$\sqrt{\Delta}=\sqrt{736}=\sqrt{16*46}=\sqrt{16}*\sqrt{46}=4\sqrt{46}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{46}}{2*3}=\frac{20-4\sqrt{46}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{46}}{2*3}=\frac{20+4\sqrt{46}}{6}
$