Solve by Completing the Square y^2+4y-3=0

$y^{2} + 4 y - 3 = 0$

Step 1

Add $3$ to both sides of the equation.

$y^{2} + 4 y = 3$

Step 2

To create a trinomial square on the left side of the equation, find a value that is equal to the square of half of $b$ .

${(\frac{b}{2})}^{2} = {(2)}^{2}$

Step 3

Add the term to each side of the equation.

$y^{2} + 4 y + {(2)}^{2} = 3 + {(2)}^{2}$

Step 4

Simplify the equation.

Simplify the left side.

Raise $2$ to the power of $2$ .

$y^{2} + 4 y + 4 = 3 + {(2)}^{2}$

Simplify the right side.

Simplify $3 + {(2)}^{2}$ .

Raise $2$ to the power of $2$ .

$y^{2} + 4 y + 4 = 3 + 4$

Add $3$ and $4$ .

$y^{2} + 4 y + 4 = 7$

Step 5

Factor the perfect trinomial square into ${(y + 2)}^{2}$ .

${(y + 2)}^{2} = 7$

Step 6

Solve the equation for $y$ .

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$y + 2 = \pm \sqrt{7}$

Subtract $2$ from both sides of the equation.

$y = \pm \sqrt{7} - 2$

Step 7

The result can be shown in multiple forms.

Exact Form:

$y = \pm \sqrt{7} - 2$

Decimal Form:

$y = 0.64575131 \dots, - 4.64575131 \dots$