Solve Using the Quadratic Formula 3x^2=17x+6

$3 x^{2} = 17 x + 6$

Step 1

Move all the expressions to the left side of the equation.

Subtract $17 x$ from both sides of the equation.

$3 x^{2} - 17 x = 6$

Subtract $6$ from both sides of the equation.

$3 x^{2} - 17 x - 6 = 0$

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 $a = 3$ , $b = - 17$ , and $c = - 6$ into the quadratic formula and solve for $x$ .

$\frac{17 \pm \sqrt{{(- 17)}^{2} - 4 \cdot (3 \cdot - 6)}}{2 \cdot 3}$

Step 4

Simplify.

Simplify the numerator.

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

$x = \frac{17 \pm \sqrt{289 - 4 \cdot 3 \cdot - 6}}{2 \cdot 3}$

Multiply $- 4 \cdot 3 \cdot - 6$ .

Multiply $- 4$ by $3$ .

$x = \frac{17 \pm \sqrt{289 - 12 \cdot - 6}}{2 \cdot 3}$

Multiply $- 12$ by $- 6$ .

$x = \frac{17 \pm \sqrt{289 + 72}}{2 \cdot 3}$

Add $289$ and $72$ .

$x = \frac{17 \pm \sqrt{361}}{2 \cdot 3}$

Rewrite $361$ as $19^{2}$ .

$x = \frac{17 \pm \sqrt{19^{2}}}{2 \cdot 3}$

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

$x = \frac{17 \pm 19}{2 \cdot 3}$

Multiply $2$ by $3$ .

$x = \frac{17 \pm 19}{6}$

Step 5

The final answer is the combination of both solutions.

$x = 6, - \frac{1}{3}$