x(3x+19)=x+3(3x+4)

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

x(3x+19)=x+3(3x+4)

We move all terms to the left:

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

We multiply parentheses

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


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

x+3(3x+4)

We multiply parentheses

x+9x+12

We add all the numbers together, and all the variables

10x+12

Back to the equation:

-(10x+12)


We get rid of parentheses

3x^2+19x-10x-12=0

We add all the numbers together, and all the variables

3x^2+9x-12=0

a = 3; b = 9; c = -12;
Δ = b2-4ac
Δ = 92-4·3·(-12)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-15}{2*3}=\frac{-24}{6}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+15}{2*3}=\frac{6}{6}
=1
$