3x^2+2x+16=7x^2+2x

Solution for 3x^2+2x+16=7x^2+2x equation:

3x^2+2x+16=7x^2+2x

We move all terms to the left:

3x^2+2x+16-(7x^2+2x)=0

We get rid of parentheses

3x^2-7x^2+2x-2x+16=0

We add all the numbers together, and all the variables

-4x^2+16=0

a = -4; b = 0; c = +16;
Δ = b2-4ac
Δ = 02-4·(-4)·16
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16}{2*-4}=\frac{-16}{-8}
=+2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16}{2*-4}=\frac{16}{-8}
=-2
$