4(x-3)+16=3(x+4)(x-3)

Solution for 4(x-3)+16=3(x+4)(x-3) equation:

4(x-3)+16=3(x+4)(x-3)

We move all terms to the left:

4(x-3)+16-(3(x+4)(x-3))=0

We multiply parentheses

4x-(3(x+4)(x-3))-12+16=0

We multiply parentheses ..

-(3(+x^2-3x+4x-12))+4x-12+16=0


We calculate terms in parentheses: -(3(+x^2-3x+4x-12)), so:

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

We multiply parentheses

3x^2-9x+12x-36

We add all the numbers together, and all the variables

3x^2+3x-36

Back to the equation:

-(3x^2+3x-36)


We add all the numbers together, and all the variables

4x-(3x^2+3x-36)+4=0

We get rid of parentheses

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

We add all the numbers together, and all the variables

-3x^2+x+40=0

a = -3; b = 1; c = +40;
Δ = b2-4ac
Δ = 12-4·(-3)·40
Δ = 481
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{-(1)-\sqrt{481}}{2*-3}=\frac{-1-\sqrt{481}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{481}}{2*-3}=\frac{-1+\sqrt{481}}{-6}
$