Solve Using the Quadratic Formula 3x=0.5x^2

$3 x = 0.5 x^{2}$

Step 1

Subtract $0.5 x^{2}$ from both sides of the equation.

$3 x - 0.5 x^{2} = 0$

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 $a = - 0.5$ , $b = 3$ , and $c = 0$ into the quadratic formula and solve for $x$ .

$\frac{- 3 \pm \sqrt{3^{2} - 4 \cdot (- 0.5 \cdot 0)}}{2 \cdot - 0.5}$

Step 4

Simplify.

Simplify the numerator.

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

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot - 0.5 \cdot 0}}{2 \cdot - 0.5}$

Multiply $- 4 \cdot - 0.5 \cdot 0$ .

Multiply $- 4$ by $- 0.5$ .

$x = \frac{- 3 \pm \sqrt{9 + 2 \cdot 0}}{2 \cdot - 0.5}$

Multiply $2$ by $0$ .

$x = \frac{- 3 \pm \sqrt{9 + 0}}{2 \cdot - 0.5}$

Add $9$ and $0$ .

$x = \frac{- 3 \pm \sqrt{9}}{2 \cdot - 0.5}$

Rewrite $9$ as $3^{2}$ .

$x = \frac{- 3 \pm \sqrt{3^{2}}}{2 \cdot - 0.5}$

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

$x = \frac{- 3 \pm 3}{2 \cdot - 0.5}$

Multiply $2$ by $- 0.5$ .

$x = \frac{- 3 \pm 3}{- 1}$

Move the negative one from the denominator of $\frac{- 3 \pm 3}{- 1}$ .

$x = - 1 \cdot (- 3 \pm 3)$

Rewrite $- 1 \cdot (- 3 \pm 3)$ as $- (- 3 \pm 3)$ .

$x = - (- 3 \pm 3)$

Step 5

The final answer is the combination of both solutions.

$x = 0, 6$