# Solve the equation (x+10)^(2)=(5-x)^(2)

Solve the equation $\left(x+10{\right)}^{2}=\left(5-x{\right)}^{2}$
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coccusk7
$\left(х+10{\right)}^{2}=\left(5-х{\right)}^{2}$
${x}^{2}+20x+100=25-10x+{x}^{2}$
${x}^{2}-{x}^{2}+20x+10x=25-100$
30x= -75
x= -75/30
x= -2,5

Andrejkoxg
$\left(x+10{\right)}^{2}=\left(5-x{\right)}^{2}$
let's square
formula:
$\left(a+b{\right)}^{2}={a}^{2}+2\ast a\ast b+{b}^{2}$
or
$\left(a-b{\right)}^{2}={a}^{2}-2\ast a\ast b+{b}^{2}$
${x}^{2}+20x+100=25-10x+{x}^{2}$
Let's move everything to one side. remember that when transferring, the sign changes to the opposite.
X is reduced and we get:
30x+75=0
$x=\frac{75}{30}=2,5$