3x2+2x=2(x-4)+5x2

Solution for 3x2+2x=2(x-4)+5x2 equation:

3x^2+2x=2(x-4)+5x^2

We move all terms to the left:

3x^2+2x-(2(x-4)+5x^2)=0


We calculate terms in parentheses: -(2(x-4)+5x^2), so:

2(x-4)+5x^2

determiningTheFunctionDomain
5x^2+2(x-4)

We multiply parentheses

5x^2+2x-8

Back to the equation:

-(5x^2+2x-8)


We add all the numbers together, and all the variables

3x^2+2x-(5x^2+2x-8)=0

We get rid of parentheses

3x^2-5x^2+2x-2x+8=0

We add all the numbers together, and all the variables

-2x^2+8=0

a = -2; b = 0; c = +8;
Δ = b2-4ac
Δ = 02-4·(-2)·8
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8}{2*-2}=\frac{-8}{-4}
=+2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8}{2*-2}=\frac{8}{-4}
=-2
$