Solve Using the Quadratic Formula 4x^2-17x+4=0

$4 x^{2} - 17 x + 4 = 0$

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 $a = 4$ , $b = - 17$ , and $c = 4$ into the quadratic formula and solve for $x$ .

$\frac{17 \pm \sqrt{{(- 17)}^{2} - 4 \cdot (4 \cdot 4)}}{2 \cdot 4}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 17$ to the power of $2$ .

$x = \frac{17 \pm \sqrt{289 - 4 \cdot 4 \cdot 4}}{2 \cdot 4}$

Multiply $- 4 \cdot 4 \cdot 4$ .

Multiply $- 4$ by $4$ .

$x = \frac{17 \pm \sqrt{289 - 16 \cdot 4}}{2 \cdot 4}$

Multiply $- 16$ by $4$ .

$x = \frac{17 \pm \sqrt{289 - 64}}{2 \cdot 4}$

Subtract $64$ from $289$ .

$x = \frac{17 \pm \sqrt{225}}{2 \cdot 4}$

Rewrite $225$ as $15^{2}$ .

$x = \frac{17 \pm \sqrt{15^{2}}}{2 \cdot 4}$

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

$x = \frac{17 \pm 15}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{17 \pm 15}{8}$

Step 4

The final answer is the combination of both solutions.

$x = 4, \frac{1}{4}$