(4x-15)=(x+1)(x-2)

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Solution for (4x-15)=(x+1)(x-2) equation: 



(4x-15)=(x+1)(x-2)

We move all terms to the left:

(4x-15)-((x+1)(x-2))=0

We get rid of parentheses

4x-((x+1)(x-2))-15=0

We multiply parentheses ..

-((+x^2-2x+x-2))+4x-15=0



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

(+x^2-2x+x-2)

We get rid of parentheses

x^2-2x+x-2

We add all the numbers together, and all the variables

x^2-1x-2

Back to the equation:

-(x^2-1x-2)



We add all the numbers together, and all the variables

4x-(x^2-1x-2)-15=0

We get rid of parentheses

-x^2+4x+1x+2-15=0

We add all the numbers together, and all the variables

-1x^2+5x-13=0

a = -1; b = 5; c = -13;
Δ = b2-4ac
Δ = 52-4·(-1)·(-13)
Δ = -27
Delta is less than zero, so there is no solution for the equation

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