2x^2+3x+19=3x^2-4x

Solution for 2x^2+3x+19=3x^2-4x equation:

2x^2+3x+19=3x^2-4x

We move all terms to the left:

2x^2+3x+19-(3x^2-4x)=0

We get rid of parentheses

2x^2-3x^2+3x+4x+19=0

We add all the numbers together, and all the variables

-1x^2+7x+19=0

a = -1; b = 7; c = +19;
Δ = b2-4ac
Δ = 72-4·(-1)·19
Δ = 125
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:

$\sqrt{\Delta}=\sqrt{125}=\sqrt{25*5}=\sqrt{25}*\sqrt{5}=5\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-5\sqrt{5}}{2*-1}=\frac{-7-5\sqrt{5}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+5\sqrt{5}}{2*-1}=\frac{-7+5\sqrt{5}}{-2}
$