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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-2x^2+7x+4=0

a = -2; b = 7; c = +4;
Δ = b2-4ac
Δ = 72-4·(-2)·4
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-9}{2*-2}=\frac{-16}{-4}
=+4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+9}{2*-2}=\frac{2}{-4}
=-1/2
$