2x^2+x-38=4x+6

Solution for 2x^2+x-38=4x+6 equation:

2x^2+x-38=4x+6

We move all terms to the left:

2x^2+x-38-(4x+6)=0

We get rid of parentheses

2x^2+x-4x-6-38=0

We add all the numbers together, and all the variables

2x^2-3x-44=0

a = 2; b = -3; c = -44;
Δ = b2-4ac
Δ = -32-4·2·(-44)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-19}{2*2}=\frac{-16}{4}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+19}{2*2}=\frac{22}{4}
=5+1/2
$