(x+3)=(x+4)(x+5)

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Solution for (x+3)=(x+4)(x+5) equation: 



(x+3)=(x+4)(x+5)

We move all terms to the left:

(x+3)-((x+4)(x+5))=0

We get rid of parentheses

x-((x+4)(x+5))+3=0

We multiply parentheses ..

-((+x^2+5x+4x+20))+x+3=0



We calculate terms in parentheses: -((+x^2+5x+4x+20)), so:

(+x^2+5x+4x+20)

We get rid of parentheses

x^2+5x+4x+20

We add all the numbers together, and all the variables

x^2+9x+20

Back to the equation:

-(x^2+9x+20)



We add all the numbers together, and all the variables

x-(x^2+9x+20)+3=0

We get rid of parentheses

-x^2+x-9x-20+3=0

We add all the numbers together, and all the variables

-1x^2-8x-17=0

a = -1; b = -8; c = -17;
Δ = b2-4ac
Δ = -82-4·(-1)·(-17)
Δ = -4
Delta is less than zero, so there is no solution for the equation

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