2x(x+3)=19(x+3)

Solution for 2x(x+3)=19(x+3) equation:

2x(x+3)=19(x+3)

We move all terms to the left:

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

We multiply parentheses

2x^2+6x-(19(x+3))=0


We calculate terms in parentheses: -(19(x+3)), so:

19(x+3)

We multiply parentheses

19x+57

Back to the equation:

-(19x+57)


We get rid of parentheses

2x^2+6x-19x-57=0

We add all the numbers together, and all the variables

2x^2-13x-57=0

a = 2; b = -13; c = -57;
Δ = b2-4ac
Δ = -132-4·2·(-57)
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-25}{2*2}=\frac{-12}{4}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+25}{2*2}=\frac{38}{4}
=9+1/2
$