(2x+1)2=4(x2+x+4)

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Solution for (2x+1)2=4(x2+x+4) equation: 



(2x+1)2=4(x2+x+4)

We move all terms to the left:

(2x+1)2-(4(x2+x+4))=0

We add all the numbers together, and all the variables

-(4(+x^2+x+4))+(2x+1)2=0

We multiply parentheses

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



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

4(+x^2+x+4)

We multiply parentheses

4x^2+4x+16

Back to the equation:

-(4x^2+4x+16)



We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-4x^2-14=0

a = -4; b = 0; c = -14;
Δ = b2-4ac
Δ = 02-4·(-4)·(-14)
Δ = -224
Delta is less than zero, so there is no solution for the equation

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