x^2+(x^2+34x+289)=25^2

Solution for x^2+(x^2+34x+289)=25^2 equation:

x^2+(x^2+34x+289)=25^2

We move all terms to the left:

x^2+(x^2+34x+289)-(25^2)=0

We add all the numbers together, and all the variables

x^2+(x^2+34x+289)-625=0

We get rid of parentheses

x^2+x^2+34x+289-625=0

We add all the numbers together, and all the variables

2x^2+34x-336=0

a = 2; b = 34; c = -336;
Δ = b2-4ac
Δ = 342-4·2·(-336)
Δ = 3844
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3844}=62$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-62}{2*2}=\frac{-96}{4}
=-24
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+62}{2*2}=\frac{28}{4}
=7
$