(7x-5)=(x+1)(x+1)

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

(7x-5)=(x+1)(x+1)

We move all terms to the left:

(7x-5)-((x+1)(x+1))=0

We get rid of parentheses

7x-((x+1)(x+1))-5=0

We multiply parentheses ..

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


We calculate terms in parentheses: -((+x^2+x+x+1)), so:

(+x^2+x+x+1)

We get rid of parentheses

x^2+x+x+1

We add all the numbers together, and all the variables

x^2+2x+1

Back to the equation:

-(x^2+2x+1)


We add all the numbers together, and all the variables

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

We get rid of parentheses

-x^2+7x-2x-1-5=0

We add all the numbers together, and all the variables

-1x^2+5x-6=0

a = -1; b = 5; c = -6;
Δ = b2-4ac
Δ = 52-4·(-1)·(-6)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-1}{2*-1}=\frac{-6}{-2}
=+3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+1}{2*-1}=\frac{-4}{-2}
=+2
$