Solve Using the Quadratic Formula 2w^2-28w=-98

$2 w^{2} - 28 w = - 98$

Step 1

Add $98$ to both sides of the equation.

$2 w^{2} - 28 w + 98 = 0$

Step 2

Use the quadratic formula to find the solutions.

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

Step 3

Substitute the values $a = 2$ , $b = - 28$ , and $c = 98$ into the quadratic formula and solve for $w$ .

$\frac{28 \pm \sqrt{{(- 28)}^{2} - 4 \cdot (2 \cdot 98)}}{2 \cdot 2}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 28$ to the power of $2$ .

$w = \frac{28 \pm \sqrt{784 - 4 \cdot 2 \cdot 98}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot 98$ .

Multiply $- 4$ by $2$ .

$w = \frac{28 \pm \sqrt{784 - 8 \cdot 98}}{2 \cdot 2}$

Multiply $- 8$ by $98$ .

$w = \frac{28 \pm \sqrt{784 - 784}}{2 \cdot 2}$

Subtract $784$ from $784$ .

$w = \frac{28 \pm \sqrt{0}}{2 \cdot 2}$

Rewrite $0$ as $0^{2}$ .

$w = \frac{28 \pm \sqrt{0^{2}}}{2 \cdot 2}$

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

$w = \frac{28 \pm 0}{2 \cdot 2}$

$28$ plus or minus $0$ is $28$ .

$w = \frac{28}{2 \cdot 2}$

Multiply $2$ by $2$ .

$w = \frac{28}{4}$

Divide $28$ by $4$ .

$w = 7$

Step 5

The final answer is the combination of both solutions.

$w = 7$ Double roots