3x+1=2x^2+4x

Solution for 3x+1=2x^2+4x equation:

3x+1=2x^2+4x

We move all terms to the left:

3x+1-(2x^2+4x)=0

We get rid of parentheses

-2x^2+3x-4x+1=0

We add all the numbers together, and all the variables

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

a = -2; b = -1; c = +1;
Δ = b2-4ac
Δ = -12-4·(-2)·1
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{-(-1)-3}{2*-2}=\frac{-2}{-4}
=1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+3}{2*-2}=\frac{4}{-4}
=-1
$