How do I solve a quadratic equation by completing the square?

Make this expression into an equation by making it equal to zero.

x² + 3x + 5 = 0 then halve-double the coefficient of x, and add and subtract the square of this halved value

x² + 2*³∕₂x + (³∕₂)² + 5 - (³∕₂)² = 0 form up the square from x² + 2gx + g² = (x+g)² and simplify the other terms

(x + ³∕₂)² + ¹¹∕₄ = 0 then manipulate into the difference of two squares

(x + ³∕₂)² - (i½√11)² = 0 then factorize

(x + ³∕₂ - i½√11)(x + ³∕₂ + i½√11) = 0 then solve for each parenthesis

x = -³∕₂ + i½√11 or -³∕₂ - i½√11

x² + 3x + 5 = 0 --- {Take the middle term, half it, square it}

x² + 3x + 9/4 - 9/4 + 5 = 0

(x + 3/2)² - 9/4 + 5 = 0

(x + 3/2)² + 11/4 = 0

That puts it into vertex form. As you can see, vertex form has only one "x" in it, so we can solve for that x using simple algebra.

(x + 3/2)² + 11/4 = 0

x + 3/2 = ±√[-11/4]

x = -3/2 ± √[-11/4]

x = (-3 ± √[-11]) ÷ 2

x² + 3x + 5 is not an expression, not equation.

Do you mean x² + 3x + 5 = 0 ?

isolate the constant

x² +3x = -5

complete the square

     coefficient of the x term: 3

     divide coefficient in half: 3/2

     square the quotient: (3/2)²

     complete the square by adding (3/2)² to both sides of the equation:

x² + 3x + (3/2)² = (3/2)² - 5

(x + 3/2)² = -11/4

x + 3/2 = ±√(-11/4) = ±(√(11/4))i = ±((√11)/2)i

x = -3/2 ±((√11)/2)i

If u equate that to zero:

(x+1.5)^2 - 1.5^2 +5

Just use the quadratic formula..easier alternative