Respuesta :

JamesAlanMoses13
Take the square root of each side of the equation to setup the solution for xx
(x−5)2⋅12=±√29x-52⋅12=±29
Remove the perfect root factor x−5x-5 under the radical to solve for xx.
(x−5)=±√29x-5=±29
Remove parentheses.
x−5=±√29x-5=±29
Since −5-5 does not contain the variable to solve for, move it to the right side of the equation by adding 55 to both sides.
x=5±√29
irspow
x^2+10x-4=0  

There are a couple of ways to solve this, you can complete the square or use the quadratic formula (which is derived from the steps to completing the square).  Factoring in this case would be very difficult as there are no easy integer values to solve it with.  So we'll complete the square:

Move the constant to the other side by adding 4 to both sides...

x^2+10x=4  Now halve the linear coefficient, 10 in this case, square that value and add it to both sides...ie 10/2=5, 5^2=25, so add 25 to both sides

x^2+10x+25=29  Now the left side is a perfect square....

(x+5)^2=29  now take the square root of both sides...

x+5=±√29  finally subtract 5 from both sides...

x=-5±√29

So x=-5+√29 and -5-√29

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