0(x)=(3x-1)(3-x)+4x+19

Solution for 0(x)=(3x-1)(3-x)+4x+19 equation:

0(x)=(3x-1)(3-x)+4x+19

We move all terms to the left:

0(x)-((3x-1)(3-x)+4x+19)=0

We add all the numbers together, and all the variables

0x-((3x-1)(-1x+3)+4x+19)=0

We add all the numbers together, and all the variables

x-((3x-1)(-1x+3)+4x+19)=0

We multiply parentheses ..

-((-3x^2+9x+x-3)+4x+19)+x=0


We calculate terms in parentheses: -((-3x^2+9x+x-3)+4x+19), so:

(-3x^2+9x+x-3)+4x+19

We get rid of parentheses

-3x^2+9x+x+4x-3+19

We add all the numbers together, and all the variables

-3x^2+14x+16

Back to the equation:

-(-3x^2+14x+16)


We get rid of parentheses

3x^2-14x+x-16=0

We add all the numbers together, and all the variables

3x^2-13x-16=0

a = 3; b = -13; c = -16;
Δ = b2-4ac
Δ = -132-4·3·(-16)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-19}{2*3}=\frac{-6}{6}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+19}{2*3}=\frac{32}{6}
=5+1/3
$