Solve Using the Quadratic Formula 7x^2=9+x

$7 x^{2} = 9 + x$

Step 1

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

Subtract $9$ from both sides of the equation.

$7 x^{2} - 9 = x$

Subtract $x$ from both sides of the equation.

$7 x^{2} - 9 - x = 0$

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 $a = 7$ , $b = - 1$ , and $c = - 9$ into the quadratic formula and solve for $x$ .

$\frac{1 \pm \sqrt{{(- 1)}^{2} - 4 \cdot (7 \cdot - 9)}}{2 \cdot 7}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 1$ to the power of $2$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 7 \cdot - 9}}{2 \cdot 7}$

Multiply $- 4 \cdot 7 \cdot - 9$ .

Multiply $- 4$ by $7$ .

$x = \frac{1 \pm \sqrt{1 - 28 \cdot - 9}}{2 \cdot 7}$

Multiply $- 28$ by $- 9$ .

$x = \frac{1 \pm \sqrt{1 + 252}}{2 \cdot 7}$

Add $1$ and $252$ .

$x = \frac{1 \pm \sqrt{253}}{2 \cdot 7}$

Multiply $2$ by $7$ .

$x = \frac{1 \pm \sqrt{253}}{14}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = \frac{1 \pm \sqrt{253}}{14}$

Decimal Form:

$x = 1.20756955 \dots, - 1.06471240 \dots$