2u^2-4u+16=(u+2^2)

Solution for 2u^2-4u+16=(u+2^2) equation:

2u^2-4u+16=(u+2^2)

We move all terms to the left:

2u^2-4u+16-((u+2^2))=0


We calculate terms in parentheses: -((u+2^2)), so:

(u+2^2)

We get rid of parentheses

u+2^2

We add all the numbers together, and all the variables

u+4

Back to the equation:

-(u+4)


We get rid of parentheses

2u^2-4u-u-4+16=0

We add all the numbers together, and all the variables

2u^2-5u+12=0

a = 2; b = -5; c = +12;
Δ = b2-4ac
Δ = -52-4·2·12
Δ = -71
Delta is less than zero, so there is no solution for the equation