2x^2=-x^2+16x=12

Solution for 2x^2=-x^2+16x=12 equation:

2x^2=-x^2+16x=12

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

3x^2-16x=0

a = 3; b = -16; c = 0;
Δ = b2-4ac
Δ = -162-4·3·0
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{-(-16)-16}{2*3}=\frac{0}{6}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16}{2*3}=\frac{32}{6}
=5+1/3
$