25+x^2-8x=137-x^2-22x

Solution for 25+x^2-8x=137-x^2-22x equation:

25+x^2-8x=137-x^2-22x

We move all terms to the left:

25+x^2-8x-(137-x^2-22x)=0

We get rid of parentheses

x^2+x^2+22x-8x-137+25=0

We add all the numbers together, and all the variables

2x^2+14x-112=0

a = 2; b = 14; c = -112;
Δ = b2-4ac
Δ = 142-4·2·(-112)
Δ = 1092
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1092}=\sqrt{4*273}=\sqrt{4}*\sqrt{273}=2\sqrt{273}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{273}}{2*2}=\frac{-14-2\sqrt{273}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{273}}{2*2}=\frac{-14+2\sqrt{273}}{4}
$