Solve for x 8x+10x^2-6=17x

$8 x + 10 x^{2} - 6 = 17 x$

Step 1

Move all terms containing $x$ to the left side of the equation.

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

$8 x + 10 x^{2} - 6 - 17 x = 0$

Subtract $17 x$ from $8 x$ .

$- 9 x + 10 x^{2} - 6 = 0$

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 $a = 10$ , $b = - 9$ , and $c = - 6$ into the quadratic formula and solve for $x$ .

$\frac{9 \pm \sqrt{{(- 9)}^{2} - 4 \cdot (10 \cdot - 6)}}{2 \cdot 10}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 9$ to the power of $2$ .

$x = \frac{9 \pm \sqrt{81 - 4 \cdot 10 \cdot - 6}}{2 \cdot 10}$

Multiply $- 4 \cdot 10 \cdot - 6$ .

Multiply $- 4$ by $10$ .

$x = \frac{9 \pm \sqrt{81 - 40 \cdot - 6}}{2 \cdot 10}$

Multiply $- 40$ by $- 6$ .

$x = \frac{9 \pm \sqrt{81 + 240}}{2 \cdot 10}$

Add $81$ and $240$ .

$x = \frac{9 \pm \sqrt{321}}{2 \cdot 10}$

Multiply $2$ by $10$ .

$x = \frac{9 \pm \sqrt{321}}{20}$

Step 5

The final answer is the combination of both solutions.

$x = \frac{9 + \sqrt{321}}{20}, \frac{9 - \sqrt{321}}{20}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = \frac{9 + \sqrt{321}}{20}, \frac{9 - \sqrt{321}}{20}$

Decimal Form:

$x = 1.34582364 \dots, - 0.44582364 \dots$