Solve Using the Quadratic Formula 7x^2-3x=3

$7 x^{2} - 3 x = 3$

Step 1

Subtract $3$ from both sides of the equation.

$7 x^{2} - 3 x - 3 = 0$

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 $a = 7$ , $b = - 3$ , and $c = - 3$ into the quadratic formula and solve for $x$ .

$\frac{3 \pm \sqrt{{(- 3)}^{2} - 4 \cdot (7 \cdot - 3)}}{2 \cdot 7}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 3$ to the power of $2$ .

$x = \frac{3 \pm \sqrt{9 - 4 \cdot 7 \cdot - 3}}{2 \cdot 7}$

Multiply $- 4 \cdot 7 \cdot - 3$ .

Multiply $- 4$ by $7$ .

$x = \frac{3 \pm \sqrt{9 - 28 \cdot - 3}}{2 \cdot 7}$

Multiply $- 28$ by $- 3$ .

$x = \frac{3 \pm \sqrt{9 + 84}}{2 \cdot 7}$

Add $9$ and $84$ .

$x = \frac{3 \pm \sqrt{93}}{2 \cdot 7}$

Multiply $2$ by $7$ .

$x = \frac{3 \pm \sqrt{93}}{14}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = \frac{3 \pm \sqrt{93}}{14}$

Decimal Form:

$x = 0.90311791 \dots, - 0.47454648 \dots$