2(x+5)=3(x-3)x+19

Solution for 2(x+5)=3(x-3)x+19 equation:

2(x+5)=3(x-3)x+19

We move all terms to the left:

2(x+5)-(3(x-3)x+19)=0

We multiply parentheses

2x-(3(x-3)x+19)+10=0


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

3(x-3)x+19

We multiply parentheses

3x^2-9x+19

Back to the equation:

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


We get rid of parentheses

-3x^2+2x+9x-19+10=0

We add all the numbers together, and all the variables

-3x^2+11x-9=0

a = -3; b = 11; c = -9;
Δ = b2-4ac
Δ = 112-4·(-3)·(-9)
Δ = 13
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{-(11)-\sqrt{13}}{2*-3}=\frac{-11-\sqrt{13}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{13}}{2*-3}=\frac{-11+\sqrt{13}}{-6}
$