Solve Using the Quadratic Formula y^2-9y+14=0

$y^{2} - 9 y + 14 = 0$

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 $a = 1$ , $b = - 9$ , and $c = 14$ into the quadratic formula and solve for $y$ .

$\frac{9 \pm \sqrt{{(- 9)}^{2} - 4 \cdot (1 \cdot 14)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 9$ to the power of $2$ .

$y = \frac{9 \pm \sqrt{81 - 4 \cdot 1 \cdot 14}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 14$ .

Multiply $- 4$ by $1$ .

$y = \frac{9 \pm \sqrt{81 - 4 \cdot 14}}{2 \cdot 1}$

Multiply $- 4$ by $14$ .

$y = \frac{9 \pm \sqrt{81 - 56}}{2 \cdot 1}$

Subtract $56$ from $81$ .

$y = \frac{9 \pm \sqrt{25}}{2 \cdot 1}$

Rewrite $25$ as $5^{2}$ .

$y = \frac{9 \pm \sqrt{5^{2}}}{2 \cdot 1}$

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

$y = \frac{9 \pm 5}{2 \cdot 1}$

Multiply $2$ by $1$ .

$y = \frac{9 \pm 5}{2}$

Step 4

The final answer is the combination of both solutions.

$y = 7, 2$