Solve Using the Quadratic Formula 4r^2=r+3

$4 r^{2} = r + 3$

Step 1

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

Subtract $r$ from both sides of the equation.

$4 r^{2} - r = 3$

Subtract $3$ from both sides of the equation.

$4 r^{2} - r - 3 = 0$

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 $a = 4$ , $b = - 1$ , and $c = - 3$ into the quadratic formula and solve for $r$ .

$\frac{1 \pm \sqrt{{(- 1)}^{2} - 4 \cdot (4 \cdot - 3)}}{2 \cdot 4}$

Step 4

Simplify.

Simplify the numerator.

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

$r = \frac{1 \pm \sqrt{1 - 4 \cdot 4 \cdot - 3}}{2 \cdot 4}$

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

Multiply $- 4$ by $4$ .

$r = \frac{1 \pm \sqrt{1 - 16 \cdot - 3}}{2 \cdot 4}$

Multiply $- 16$ by $- 3$ .

$r = \frac{1 \pm \sqrt{1 + 48}}{2 \cdot 4}$

Add $1$ and $48$ .

$r = \frac{1 \pm \sqrt{49}}{2 \cdot 4}$

Rewrite $49$ as $7^{2}$ .

$r = \frac{1 \pm \sqrt{7^{2}}}{2 \cdot 4}$

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

$r = \frac{1 \pm 7}{2 \cdot 4}$

Multiply $2$ by $4$ .

$r = \frac{1 \pm 7}{8}$

Step 5

The final answer is the combination of both solutions.

$r = 1, - \frac{3}{4}$