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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

x^2-10x=0

a = 1; b = -10; c = 0;
Δ = b2-4ac
Δ = -102-4·1·0
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-10}{2*1}=\frac{0}{2}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+10}{2*1}=\frac{20}{2}
=10
$