Precalculus Examples

$7 x = 2 x^{2} + 1$

Step 1

Subtract $2 x^{2}$ from both sides of the equation.

$7 x - 2 x^{2} = 1$

Step 2

Subtract $1$ from both sides of the equation.

$7 x - 2 x^{2} - 1 = 0$

Step 3

Use the quadratic formula to find the solutions.

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

Step 4

Substitute the values $a = - 2$ , $b = 7$ , and $c = - 1$ into the quadratic formula and solve for $x$ .

$\frac{- 7 \pm \sqrt{7^{2} - 4 \cdot (- 2 \cdot - 1)}}{2 \cdot - 2}$

Step 5

Simplify.

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Step 5.1

Simplify the numerator.

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Step 5.1.1

Raise $7$ to the power of $2$ .

$x = \frac{- 7 \pm \sqrt{49 - 4 \cdot - 2 \cdot - 1}}{2 \cdot - 2}$

Step 5.1.2

Multiply $- 4 \cdot - 2 \cdot - 1$ .

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Step 5.1.2.1

Multiply $- 4$ by $- 2$ .

$x = \frac{- 7 \pm \sqrt{49 + 8 \cdot - 1}}{2 \cdot - 2}$

Step 5.1.2.2

Multiply $8$ by $- 1$ .

$x = \frac{- 7 \pm \sqrt{49 - 8}}{2 \cdot - 2}$

Step 5.1.3

Subtract $8$ from $49$ .

$x = \frac{- 7 \pm \sqrt{41}}{2 \cdot - 2}$

Step 5.2

Multiply $2$ by $- 2$ .

$x = \frac{- 7 \pm \sqrt{41}}{- 4}$

Step 5.3

Simplify $\frac{- 7 \pm \sqrt{41}}{- 4}$ .

$x = \frac{7 \pm \sqrt{41}}{4}$

Step 6

The final answer is the combination of both solutions.

$x = \frac{7 + \sqrt{41}}{4}, \frac{7 - \sqrt{41}}{4}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$x = \frac{7 + \sqrt{41}}{4}, \frac{7 - \sqrt{41}}{4}$

Decimal Form:

$x = 3.35078105 \dots, 0.14921894 \dots$