Verified Answer :

xSamx

You take half the coefficient of the x term and then square it and then add it to both sides.
\(x^2- 10x = 7\)
\((x^2-10x +25)=7+25[\)\((x^2-10x +25)=32\) 
\((x-5)^2=32\)
In your case, 25 should be added to both sides complete the square

calculista

we have

\(x^{2} -10x=7\)

Complete the square. Remember to balance the equation by adding the same constants to each side

to complete the square adds \((10/2)^{2}=25\) both sides

so

\(x^{2} -10x+25=7+25\)

\(x^{2} -10x+25=32\)

Rewrite as perfect squares

\((x-5)^{2}=32\)

Square root both sides

\(x-5=(+/-)\sqrt{32}\)

\(x=5(+/-)\sqrt{32}\)

\(x1=5+\sqrt{32}\)

\(x2=5-\sqrt{32}\)

therefore

the answer is

The number is \(25\)




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