On solving the quadratic equation by completing the square root, we get
[tex]-32=2(x^2+10)\\-32+50=2(x^2+10x+25)\\18=2(x+5)^2\\9=(x+5)^2\\\pm {3}=x+5\\x=-2\text{ or }x=-8[/tex]
The value of the variable in the quadratic polynomial is said to be the solution if the value of the function is zero.
We will use the identity, [tex](a+b)^2=a^2+2ab+b^2[/tex] to determine the missing terms.
In the equation, [tex]-32+?=2(x^2+10x+25)[/tex]; [tex]50[/tex] is added to the right-hand side of the equation so, to balance the equation, add [tex]50[/tex].
So, [tex]-32+50=2(x^2+10x+25)[/tex].
Now, take the square root to both sides of the equation [tex]9=(x+5)^2[/tex] as-
[tex]\sqrt{9}=\sqrt{(x+5)^2}\\\pm {3}=x+5[/tex]
Now, solve the equation [tex]\pm {3}=x+5[/tex] as-
[tex]x+5=3\text{ and } x+5=-3\\x=-2\text{ and } x=-8[/tex]
Thus, on solving the quadratic equation by completing the square root, we get
[tex]-32=2(x^2+10)\\-32+50=2(x^2+10x+25)\\18=2(x+5)^2\\9=(x+5)^2\\\pm {3}=x+5\\x=-2\text{ or }x=-8[/tex]
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