12-(2x-3)(-x+4)=7x-(3-x)x

Solution for 12-(2x-3)(-x+4)=7x-(3-x)x equation:

12-(2x-3)(-x+4)=7x-(3-x)x

We move all terms to the left:

12-(2x-3)(-x+4)-(7x-(3-x)x)=0

We add all the numbers together, and all the variables

-(2x-3)(-1x+4)-(7x-(-1x+3)x)+12=0

We multiply parentheses ..

-(-2x^2+8x+3x-12)-(7x-(-1x+3)x)+12=0


We calculate terms in parentheses: -(7x-(-1x+3)x), so:

7x-(-1x+3)x

We multiply parentheses

1x^2+7x-3x

We add all the numbers together, and all the variables

x^2+4x

Back to the equation:

-(x^2+4x)


We get rid of parentheses

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

We add all the numbers together, and all the variables

x^2-15x+24=0

a = 1; b = -15; c = +24;
Δ = b2-4ac
Δ = -152-4·1·24
Δ = 129
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{-(-15)-\sqrt{129}}{2*1}=\frac{15-\sqrt{129}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{129}}{2*1}=\frac{15+\sqrt{129}}{2}
$