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

Solution for (7-x)2+(5+2x)(7-x)=0 equation:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

-2x+(2x+5)(-1x+7)+14=0

We multiply parentheses ..

(-2x^2+14x-5x+35)-2x+14=0

We get rid of parentheses

-2x^2+14x-5x-2x+35+14=0

We add all the numbers together, and all the variables

-2x^2+7x+49=0

a = -2; b = 7; c = +49;
Δ = b2-4ac
Δ = 72-4·(-2)·49
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-21}{2*-2}=\frac{-28}{-4}
=+7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+21}{2*-2}=\frac{14}{-4}
=-3+1/2
$