3x(-12-x)=7x+(10+2x)-6

Solution for 3x(-12-x)=7x+(10+2x)-6 equation:

3x(-12-x)=7x+(10+2x)-6

We move all terms to the left:

3x(-12-x)-(7x+(10+2x)-6)=0

We add all the numbers together, and all the variables

3x(-1x-12)-(7x+(2x+10)-6)=0

We multiply parentheses

-3x^2-36x-(7x+(2x+10)-6)=0


We calculate terms in parentheses: -(7x+(2x+10)-6), so:

7x+(2x+10)-6

We get rid of parentheses

7x+2x+10-6

We add all the numbers together, and all the variables

9x+4

Back to the equation:

-(9x+4)


We get rid of parentheses

-3x^2-36x-9x-4=0

We add all the numbers together, and all the variables

-3x^2-45x-4=0

a = -3; b = -45; c = -4;
Δ = b2-4ac
Δ = -452-4·(-3)·(-4)
Δ = 1977
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-45)-\sqrt{1977}}{2*-3}=\frac{45-\sqrt{1977}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-45)+\sqrt{1977}}{2*-3}=\frac{45+\sqrt{1977}}{-6}
$