Solve Using the Quadratic Formula y^2-4y-21=0

$y^{2} - 4 y - 21 = 0$

Step 1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 2

Substitute the values $a = 1$ , $b = - 4$ , and $c = - 21$ into the quadratic formula and solve for $y$ .

$\frac{4 \pm \sqrt{{(- 4)}^{2} - 4 \cdot (1 \cdot - 21)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 4$ to the power of $2$ .

$y = \frac{4 \pm \sqrt{16 - 4 \cdot 1 \cdot - 21}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 21$ .

Multiply $- 4$ by $1$ .

$y = \frac{4 \pm \sqrt{16 - 4 \cdot - 21}}{2 \cdot 1}$

Multiply $- 4$ by $- 21$ .

$y = \frac{4 \pm \sqrt{16 + 84}}{2 \cdot 1}$

Add $16$ and $84$ .

$y = \frac{4 \pm \sqrt{100}}{2 \cdot 1}$

Rewrite $100$ as $10^{2}$ .

$y = \frac{4 \pm \sqrt{10^{2}}}{2 \cdot 1}$

Pull terms out from under the radical, assuming positive real numbers.

$y = \frac{4 \pm 10}{2 \cdot 1}$

Multiply $2$ by $1$ .

$y = \frac{4 \pm 10}{2}$

Simplify $\frac{4 \pm 10}{2}$ .

$y = 2 \pm 5$

Step 4

The final answer is the combination of both solutions.

$y = 7, - 3$