7(x-3)-5(x2-1)=x2-5(x+2)

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Solution for 7(x-3)-5(x2-1)=x2-5(x+2) equation: 



7(x-3)-5(x2-1)=x2-5(x+2)

We move all terms to the left:

7(x-3)-5(x2-1)-(x2-5(x+2))=0

We add all the numbers together, and all the variables

-5(+x^2-1)+7(x-3)-(x2-5(x+2))=0

We multiply parentheses

-5x^2+7x-(x2-5(x+2))+5-21=0



We calculate terms in parentheses: -(x2-5(x+2)), so:

x2-5(x+2)

We add all the numbers together, and all the variables

x^2-5(x+2)

We multiply parentheses

x^2-5x-10

Back to the equation:

-(x^2-5x-10)



We add all the numbers together, and all the variables

-5x^2+7x-(x^2-5x-10)-16=0

We get rid of parentheses

-5x^2-x^2+7x+5x+10-16=0

We add all the numbers together, and all the variables

-6x^2+12x-6=0

a = -6; b = 12; c = -6;
Δ = b2-4ac
Δ = 122-4·(-6)·(-6)
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{-12}{-12}=1$

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