5z-7=(2z+2)(2z+2)

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Solution for 5z-7=(2z+2)(2z+2) equation: 



5z-7=(2z+2)(2z+2)

We move all terms to the left:

5z-7-((2z+2)(2z+2))=0

We multiply parentheses ..

-((+4z^2+4z+4z+4))+5z-7=0



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

(+4z^2+4z+4z+4)

We get rid of parentheses

4z^2+4z+4z+4

We add all the numbers together, and all the variables

4z^2+8z+4

Back to the equation:

-(4z^2+8z+4)



We add all the numbers together, and all the variables

5z-(4z^2+8z+4)-7=0

We get rid of parentheses

-4z^2+5z-8z-4-7=0

We add all the numbers together, and all the variables

-4z^2-3z-11=0

a = -4; b = -3; c = -11;
Δ = b2-4ac
Δ = -32-4·(-4)·(-11)
Δ = -167
Delta is less than zero, so there is no solution for the equation

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