# solve by completing the square?

x^2+10x+5=0

Given  that equation is   x2+10x+5 =0

Now solve the equation using factor method.

Quadratic formula: x = [-b ± √(b2- 4ac)]/2a

Compare equation with standard from ax2+bx+c=0 and write the coefficients.  a=1, b=10 and c=5

Substitute a=1, b=10 and c=5 in the quadratic formula.

= [ -10 ± √((10)2- 4(1)(5))]/2(1)

= [-10 ± √(100 -20)]/2

= (-10 ± √(80)/2

= [(-10 ±√(16(5))]/2                              √16 = √(4)(4)=4

= -10 ± [4 √(5)]/2

= -5 ± 2 √(5)

Therefore the values of x = -5+2√(5)   (or)   x = -5-2√(5).

+1 vote
x^2 + 10x + 5 = 0
[x+5]^2-20=0
[x+5]^2=20
x+5=(+or-)sqrt20
x=-5(+or-)sqrt20

The equation is x2 + 10x + 5 = 0

Separate variables and constants aside by subtracting 5 to each side.

x2 + 10x = - 5

To change the expression into a perfect square trinomial add (half the x coefficient)² to each side of the expression

Here x coefficient = 10. so, (half the x coefficient)² = (10/2)2= 25

x2 + 10x + 25 = - 5 + 25

(x + 5)2= 20

Take square root both sides

x + 5 = ± √20

x + 5 = ± 2√5

x = - 5 ± 2√5

x = - 5 - 2√5  and  x = - 5 + 2√5.