Find All Complex Solutions 36x^2-12x+1=4

$36 x^{2} - 12 x + 1 = 4$

Step 1

Move all terms to the left side of the equation and simplify.

Subtract $4$ from both sides of the equation.

$36 x^{2} - 12 x + 1 - 4 = 0$

Subtract $4$ from $1$ .

$36 x^{2} - 12 x - 3 = 0$

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 $a = 36$ , $b = - 12$ , and $c = - 3$ into the quadratic formula and solve for $x$ .

$\frac{12 \pm \sqrt{{(- 12)}^{2} - 4 \cdot (36 \cdot - 3)}}{2 \cdot 36}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 12$ to the power of $2$ .

$x = \frac{12 \pm \sqrt{144 - 4 \cdot 36 \cdot - 3}}{2 \cdot 36}$

Multiply $- 4 \cdot 36 \cdot - 3$ .

Multiply $- 4$ by $36$ .

$x = \frac{12 \pm \sqrt{144 - 144 \cdot - 3}}{2 \cdot 36}$

Multiply $- 144$ by $- 3$ .

$x = \frac{12 \pm \sqrt{144 + 432}}{2 \cdot 36}$

Add $144$ and $432$ .

$x = \frac{12 \pm \sqrt{576}}{2 \cdot 36}$

Rewrite $576$ as $24^{2}$ .

$x = \frac{12 \pm \sqrt{24^{2}}}{2 \cdot 36}$

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

$x = \frac{12 \pm 24}{2 \cdot 36}$

Multiply $2$ by $36$ .

$x = \frac{12 \pm 24}{72}$

Simplify $\frac{12 \pm 24}{72}$ .

$x = \frac{1 \pm 2}{6}$

Step 5

The final answer is the combination of both solutions.

$x = \frac{1}{2}, - \frac{1}{6}$