Solve for b 4b^2-19b+17=0

$4 b^{2} - 19 b + 17 = 0$

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 $a = 4$ , $b = - 19$ , and $c = 17$ into the quadratic formula and solve for $b$ .

$\frac{19 \pm \sqrt{{(- 19)}^{2} - 4 \cdot (4 \cdot 17)}}{2 \cdot 4}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 19$ to the power of $2$ .

$b = \frac{19 \pm \sqrt{361 - 4 \cdot 4 \cdot 17}}{2 \cdot 4}$

Multiply $- 4 \cdot 4 \cdot 17$ .

Multiply $- 4$ by $4$ .

$b = \frac{19 \pm \sqrt{361 - 16 \cdot 17}}{2 \cdot 4}$

Multiply $- 16$ by $17$ .

$b = \frac{19 \pm \sqrt{361 - 272}}{2 \cdot 4}$

Subtract $272$ from $361$ .

$b = \frac{19 \pm \sqrt{89}}{2 \cdot 4}$

Multiply $2$ by $4$ .

$b = \frac{19 \pm \sqrt{89}}{8}$

Step 4

The final answer is the combination of both solutions.

$b = \frac{19 + \sqrt{89}}{8}, \frac{19 - \sqrt{89}}{8}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$b = \frac{19 + \sqrt{89}}{8}, \frac{19 - \sqrt{89}}{8}$

Decimal Form:

$b = 3.55424764 \dots, 1.19575235 \dots$