Fill in the Blank 3 points Identify the two numbers that multiply to 28 and add to -16 square and square

To solve this, we need to find two numbers that multiply to 28 and add to -16.<br /><br />### Step 1: Set up the equations<br />Let the two numbers be \( x \) and \( y \). Then:<br />\[<br />x \cdot y = 28 \quad \text{(the product is 28)}<br />\]<br />\[<br />x + y = -16 \quad \text{(the sum is -16)}.<br />\]<br /><br />### Step 2: Solve for one variable<br />From the second equation, solve for \( y \) in terms of \( x \):<br />\[<br />y = -16 - x.<br />\]<br /><br />### Step 3: Substitute into the first equation<br />Substitute \( y = -16 - x \) into \( x \cdot y = 28 \):<br />\[<br />x \cdot (-16 - x) = 28.<br />\]<br />Simplify:<br />\[<br />-16x - x^2 = 28.<br />\]<br />Rearrange into standard quadratic form:<br />\[<br />x^2 + 16x + 28 = 0.<br />\]<br /><br />### Step 4: Solve the quadratic equation<br />Use the quadratic formula:<br />\[<br />x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},<br />\]<br />where \( a = 1 \), \( b = 16 \), and \( c = 28 \).<br /><br />Substitute these values:<br />\[<br />x = \frac{-16 \pm \sqrt{16^2 - 4(1)(28)}}{2(1)}.<br />\]<br />\[<br />x = \frac{-16 \pm \sqrt{256 - 112}}{2}.<br />\]<br />\[<br />x = \frac{-16 \pm \sqrt{144}}{2}.<br />\]<br />\[<br />x = \frac{-16 \pm 12}{2}.<br />\]<br /><br />### Step 5: Find the two solutions<br />Solve for the two possible values of \( x \):<br />1. \( x = \frac{-16 + 12}{2} = \frac{-4}{2} = -2 \),<br />2. \( x = \frac{-16 - 12}{2} = \frac{-28}{2} = -14 \).<br /><br />Thus, the two numbers are **-2** and **-14**.<br /><br />### Final Answer:<br />The two numbers are **-2** and **-14**.