Find the Roots (Zeros) f(x)=4x^2-17x+3

$f (x) = 4 x^{2} - 17 x + 3$

Step 1

Set $4 x^{2} - 17 x + 3$ equal to $0$ .

$4 x^{2} - 17 x + 3 = 0$

Step 2

Solve for $x$ .

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 4$ , $b = - 17$ , and $c = 3$ into the quadratic formula and solve for $x$ .

$\frac{17 \pm \sqrt{{(- 17)}^{2} - 4 \cdot (4 \cdot 3)}}{2 \cdot 4}$

Simplify.

Simplify the numerator.

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

$x = \frac{17 \pm \sqrt{289 - 4 \cdot 4 \cdot 3}}{2 \cdot 4}$

Multiply $- 4 \cdot 4 \cdot 3$ .

Multiply $- 4$ by $4$ .

$x = \frac{17 \pm \sqrt{289 - 16 \cdot 3}}{2 \cdot 4}$

Multiply $- 16$ by $3$ .

$x = \frac{17 \pm \sqrt{289 - 48}}{2 \cdot 4}$

Subtract $48$ from $289$ .

$x = \frac{17 \pm \sqrt{241}}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{17 \pm \sqrt{241}}{8}$

The final answer is the combination of both solutions.

$x = \frac{17 + \sqrt{241}}{8}, \frac{17 - \sqrt{241}}{8}$

$x = \frac{17 \pm \sqrt{241}}{8}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$x = \frac{17 \pm \sqrt{241}}{8}$

Decimal Form:

$x = 4.06552183 \dots, 0.18447816 \dots$

Step 4