3x(x+4)-8=6x+10

Solution for 3x(x+4)-8=6x+10 equation:

3x(x+4)-8=6x+10

We move all terms to the left:

3x(x+4)-8-(6x+10)=0

We multiply parentheses

3x^2+12x-(6x+10)-8=0

We get rid of parentheses

3x^2+12x-6x-10-8=0

We add all the numbers together, and all the variables

3x^2+6x-18=0

a = 3; b = 6; c = -18;
Δ = b2-4ac
Δ = 62-4·3·(-18)
Δ = 252
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:

$\sqrt{\Delta}=\sqrt{252}=\sqrt{36*7}=\sqrt{36}*\sqrt{7}=6\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{7}}{2*3}=\frac{-6-6\sqrt{7}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{7}}{2*3}=\frac{-6+6\sqrt{7}}{6}
$