3x^2+18x+29=2x^2+11x+19

Solution for 3x^2+18x+29=2x^2+11x+19 equation:

3x^2+18x+29=2x^2+11x+19

We move all terms to the left:

3x^2+18x+29-(2x^2+11x+19)=0

We get rid of parentheses

3x^2-2x^2+18x-11x-19+29=0

We add all the numbers together, and all the variables

x^2+7x+10=0

a = 1; b = 7; c = +10;
Δ = b2-4ac
Δ = 72-4·1·10
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{-(7)-3}{2*1}=\frac{-10}{2}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+3}{2*1}=\frac{-4}{2}
=-2
$