Solve Using the Quadratic Formula 7k^2-6k+3=3

$7 k^{2} - 6 k + 3 = 3$

Step 1

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

Subtract $3$ from both sides of the equation.

$7 k^{2} - 6 k + 3 - 3 = 0$

Combine the opposite terms in $7 k^{2} - 6 k + 3 - 3$ .

Subtract $3$ from $3$ .

$7 k^{2} - 6 k + 0 = 0$

Add $7 k^{2} - 6 k$ and $0$ .

$7 k^{2} - 6 k = 0$

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 $a = 7$ , $b = - 6$ , and $c = 0$ into the quadratic formula and solve for $k$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (7 \cdot 0)}}{2 \cdot 7}$

Step 4

Simplify.

Simplify the numerator.

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

$k = \frac{6 \pm \sqrt{36 - 4 \cdot 7 \cdot 0}}{2 \cdot 7}$

Multiply $- 4 \cdot 7 \cdot 0$ .

Multiply $- 4$ by $7$ .

$k = \frac{6 \pm \sqrt{36 - 28 \cdot 0}}{2 \cdot 7}$

Multiply $- 28$ by $0$ .

$k = \frac{6 \pm \sqrt{36 + 0}}{2 \cdot 7}$

Add $36$ and $0$ .

$k = \frac{6 \pm \sqrt{36}}{2 \cdot 7}$

Rewrite $36$ as $6^{2}$ .

$k = \frac{6 \pm \sqrt{6^{2}}}{2 \cdot 7}$

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

$k = \frac{6 \pm 6}{2 \cdot 7}$

Multiply $2$ by $7$ .

$k = \frac{6 \pm 6}{14}$

Simplify $\frac{6 \pm 6}{14}$ .

$k = \frac{3 \pm 3}{7}$

Step 5

The final answer is the combination of both solutions.

$k = \frac{6}{7}, 0$