Solve Using the Quadratic Formula 9x^2=1-4x

$9 x^{2} = 1 - 4 x$

Step 1

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

Subtract $1$ from both sides of the equation.

$9 x^{2} - 1 = - 4 x$

Add $4 x$ to both sides of the equation.

$9 x^{2} - 1 + 4 x = 0$

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 $a = 9$ , $b = 4$ , and $c = - 1$ into the quadratic formula and solve for $x$ .

$\frac{- 4 \pm \sqrt{4^{2} - 4 \cdot (9 \cdot - 1)}}{2 \cdot 9}$

Step 4

Simplify.

Simplify the numerator.

Raise $4$ to the power of $2$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 9 \cdot - 1}}{2 \cdot 9}$

Multiply $- 4 \cdot 9 \cdot - 1$ .

Multiply $- 4$ by $9$ .

$x = \frac{- 4 \pm \sqrt{16 - 36 \cdot - 1}}{2 \cdot 9}$

Multiply $- 36$ by $- 1$ .

$x = \frac{- 4 \pm \sqrt{16 + 36}}{2 \cdot 9}$

Add $16$ and $36$ .

$x = \frac{- 4 \pm \sqrt{52}}{2 \cdot 9}$

Rewrite $52$ as $2^{2} \cdot 13$ .

Factor $4$ out of $52$ .

$x = \frac{- 4 \pm \sqrt{4 (13)}}{2 \cdot 9}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 4 \pm \sqrt{2^{2} \cdot 13}}{2 \cdot 9}$

Pull terms out from under the radical.

$x = \frac{- 4 \pm 2 \sqrt{13}}{2 \cdot 9}$

Multiply $2$ by $9$ .

$x = \frac{- 4 \pm 2 \sqrt{13}}{18}$

Simplify $\frac{- 4 \pm 2 \sqrt{13}}{18}$ .

$x = \frac{- 2 \pm \sqrt{13}}{9}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = \frac{- 2 \pm \sqrt{13}}{9}$

Decimal Form:

$x = 0.17839458 \dots, - 0.62283903 \dots$