x^2+19x+12=2(3x-14)

Solution for x^2+19x+12=2(3x-14) equation:

x^2+19x+12=2(3x-14)

We move all terms to the left:

x^2+19x+12-(2(3x-14))=0


We calculate terms in parentheses: -(2(3x-14)), so:

2(3x-14)

We multiply parentheses

6x-28

Back to the equation:

-(6x-28)


We get rid of parentheses

x^2+19x-6x+28+12=0

We add all the numbers together, and all the variables

x^2+13x+40=0

a = 1; b = 13; c = +40;
Δ = b2-4ac
Δ = 132-4·1·40
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-3}{2*1}=\frac{-16}{2}
=-8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+3}{2*1}=\frac{-10}{2}
=-5
$