Solve Using the Quadratic Formula 3x^2=6x-3

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

Step 1

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

Subtract $6 x$ from both sides of the equation.

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

Add $3$ to both sides of the equation.

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

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

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

Step 4

Simplify.

Simplify the numerator.

Raise $- 6$ to the power of $2$ .

$x = \frac{6 \pm \sqrt{36 - 4 \cdot 3 \cdot 3}}{2 \cdot 3}$

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

Multiply $- 4$ by $3$ .

$x = \frac{6 \pm \sqrt{36 - 12 \cdot 3}}{2 \cdot 3}$

Multiply $- 12$ by $3$ .

$x = \frac{6 \pm \sqrt{36 - 36}}{2 \cdot 3}$

Subtract $36$ from $36$ .

$x = \frac{6 \pm \sqrt{0}}{2 \cdot 3}$

Rewrite $0$ as $0^{2}$ .

$x = \frac{6 \pm \sqrt{0^{2}}}{2 \cdot 3}$

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

$x = \frac{6 \pm 0}{2 \cdot 3}$

$6$ plus or minus $0$ is $6$ .

$x = \frac{6}{2 \cdot 3}$

Multiply $2$ by $3$ .

$x = \frac{6}{6}$

Divide $6$ by $6$ .

$x = 1$

Step 5

The final answer is the combination of both solutions.

$x = 1$ Double roots