(2x+5)(2x+5)=(7-x)(7-x)

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

(2x+5)(2x+5)=(7-x)(7-x)

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+4x^2+10x+10x+25)-((-1x+7)(-1x+7))=0


We calculate terms in parentheses: -((-1x+7)(-1x+7)), so:

(-1x+7)(-1x+7)

We multiply parentheses ..

(+x^2-7x-7x+49)

We get rid of parentheses

x^2-7x-7x+49

We add all the numbers together, and all the variables

x^2-14x+49

Back to the equation:

-(x^2-14x+49)


We get rid of parentheses

4x^2-x^2+10x+10x+14x+25-49=0

We add all the numbers together, and all the variables

3x^2+34x-24=0

a = 3; b = 34; c = -24;
Δ = b2-4ac
Δ = 342-4·3·(-24)
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1444}=38$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-38}{2*3}=\frac{-72}{6}
=-12
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+38}{2*3}=\frac{4}{6}
=2/3
$