How do you solve using the completing the square method $x^2+10x-4=0$ ?

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Answer 1 · 2016-08-01T10:37:14

$x=+-sqrt29-5$

You can express your quadratic equation as:
$x^2+10x-4=(x+a)^2+b$

$(x+a)^2+b=x^2+2ax+a^2+b$

You will find $2ax=10x$ hence $a=5$
$a^2+b=-4$
$5^2+b=-4$
$b=-29$

Now you have: $(x+5)^2-29=0$
$(x+5)^2=29$
$(x+5)=+-sqrt29$
$thereforex=+-sqrt29-5$

Answer 2 · 2016-08-01T10:43:15

$x= -5+-sqrt29$

$x^2+10x-4=0 or (x+5)^2 -25-4=0 or (x+5)^2=29 or (x+5)=+-sqrt29 or x= -5+-sqrt29$ [Ans]

How do you solve using the completing the square method x^2+10x-4=0x^2+10x-4=0?