x+7(3x+19)=(x)(3x+4)

Solution for x+7(3x+19)=(x)(3x+4) equation:

x+7(3x+19)=(x)(3x+4)

We move all terms to the left:

x+7(3x+19)-((x)(3x+4))=0

We multiply parentheses

x+21x-(x(3x+4))+133=0


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

x(3x+4)

We multiply parentheses

3x^2+4x

Back to the equation:

-(3x^2+4x)


We add all the numbers together, and all the variables

22x-(3x^2+4x)+133=0

We get rid of parentheses

-3x^2+22x-4x+133=0

We add all the numbers together, and all the variables

-3x^2+18x+133=0

a = -3; b = 18; c = +133;
Δ = b2-4ac
Δ = 182-4·(-3)·133
Δ = 1920
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1920}=\sqrt{64*30}=\sqrt{64}*\sqrt{30}=8\sqrt{30}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-8\sqrt{30}}{2*-3}=\frac{-18-8\sqrt{30}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+8\sqrt{30}}{2*-3}=\frac{-18+8\sqrt{30}}{-6}
$