(2x-1)(x+3)=(x^2+x-7)

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

(2x-1)(x+3)=(x^2+x-7)

We move all terms to the left:

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

We multiply parentheses ..

(+2x^2+6x-1x-3)-((x^2+x-7))=0


We calculate terms in parentheses: -((x^2+x-7)), so:

(x^2+x-7)

We get rid of parentheses

x^2+x-7

Back to the equation:

-(x^2+x-7)


We get rid of parentheses

2x^2-x^2+6x-1x-x-3+7=0

We add all the numbers together, and all the variables

x^2+4x+4=0

a = 1; b = 4; c = +4;
Δ = b2-4ac
Δ = 42-4·1·4
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{-4}{2}=-2$