# Solve for x: 1. \ x^2+6=15\\ 2. \ (x+4)^2=6

## Question:

Solve for x:

{eq}1. \ x^2+6=15\\ 2. \ (x+4)^2=6 {/eq}

A quadratic equation is an equation that takes the form {eq}ax^2 + bc + c = 0 {/eq} where {eq}\rm a\, b\, and \, c {/eq} are constants greater than zero. To solve for the variable in a quadratic equation, we use the quadratic formula, factorization or the graphical method.

1. {eq}x^2 + 6 = 15 {/eq}

To solve for x in the above equation, we will start by eliminating 6 in from the LHS of the equation. We will do this by subtracting 6 fro both sides of the equation.

Therefore:

• {eq}x^2 + 6 - 6 = 15 - 6 {/eq}
• {eq}x^2 = 9 {/eq}

Finding the square root on both sides of the equation:

• {eq}\sqrt{x^2} = \sqrt{9} {/eq}
• {eq}x = \pm 3 {/eq}

Thus:

• {eq}\boxed{x = 3,\quad x = -3} {/eq}

2. {eq}(x + 4)^2 = 6 {/eq}

In the above equation, we will start by expanding the LHS of the equation.

• {eq}(x + 4)(x + 4) = 6 {/eq}
• {eq}x^2 + 8x + 16 = 6 {/eq}

Subtracting 6 from both sides of the equation, we have:

• {eq}x^2 + 8x + 16 - 6 = 6 - 6 {/eq}
• {eq}x^2 + 8x + 10 = 0 {/eq}

To solve for x in this quadratic equation, we will use the quadratic formula. The quadratic formula is given by:

• {eq}x = \dfrac{-b\pm \sqrt{b^2 -4ac}}{2a} {/eq}

• {eq}a = 1,\, b = 8,\, c = 10 {/eq}

Therefore:

• {eq}x = \dfrac{-8\pm \sqrt{(-8)^2 -4\cdot 1\cdot 10}}{2\times 1} {/eq}
• {eq}x = \dfrac{-8\pm \sqrt{64 -40}}{2} {/eq}
• {eq}x = \dfrac{-8\pm 4.8990}{2} {/eq}
• {eq}\boxed{x \approx -1.55, \quad x \approx -6.45} {/eq}