How to solve 5x^2+20x+13=0 by completing the square?

2 Answers
maganbhai P.
Mar 1, 2018

x=(-10+-sqrt(35))/5

Explanation:

5x^2+20x+13=0
Multiplying both sides by 5, we get
25x^2+100x+65=0
rArr25x^2+2(5x)(10)+100-35=0
rArr(5x)^2+2(5x)(10)+(10)^2-35=0
rArr(5x+10)^2-35=0
rArr(5x+10)^2=35=(sqrt(35))^2
rArr5x+10=+-sqrt35
5x=-10+-sqrt35
x=(-10+-sqrt35)/5

Sahar Mulla ❤
Mar 2, 2018

5x^2+20x+13=0

1/5(x^2 + 4x +13/5) = 0

x^2 + 4x +13/5=0

Now, add and subtract, half of the x-term squared, i.e. (4/2)^2=2^2 = 4

=>[x^2 + 4x +4] +[13/5-4]=0

=>[x^2 + 4x +2^2] =[4-13/5]

=>(x+2)^2 =20/5-13/5

=>(x+2) =+-sqrt(7/5

=>x =-2+-sqrt(7/5

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