x(x-3)+x-4(-3x+12)=30

Solution for x(x-3)+x-4(-3x+12)=30 equation:

x(x-3)+x-4(-3x+12)=30

We move all terms to the left:

x(x-3)+x-4(-3x+12)-(30)=0

We add all the numbers together, and all the variables

x+x(x-3)-4(-3x+12)-30=0

We multiply parentheses

x^2+x-3x+12x-48-30=0

We add all the numbers together, and all the variables

x^2+10x-78=0

a = 1; b = 10; c = -78;
Δ = b2-4ac
Δ = 102-4·1·(-78)
Δ = 412
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:

$\sqrt{\Delta}=\sqrt{412}=\sqrt{4*103}=\sqrt{4}*\sqrt{103}=2\sqrt{103}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{103}}{2*1}=\frac{-10-2\sqrt{103}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{103}}{2*1}=\frac{-10+2\sqrt{103}}{2}
$