Question

solve each of the quadratic equations.
$3x = 0.5x^2$
$\circ$ $x = -6$ or $x = 0$
$\circ$ $x = -4$ or $x = 3$
$\circ$ $x = -2$ or $x = 1.5$
$\circ$ $x = 0$ or $x = 6$
done
$0 = 5x^2 - 2x + 6$
$\circ$ $x = \frac{1 \pm 3i}{2}$
$\circ$ $x = \frac{1 \pm \sqrt{11}}{5}$
$\circ$ $x = \frac{1 \pm i\sqrt{29}}{5}$
done

Explanation:

Step1: Rearrange first equation

$3x = 0.5x^2$
Rearrange to standard form:
$0.5x^2 - 3x = 0$
Multiply by 2 to eliminate decimals:
$x^2 - 6x = 0$

Step2: Factor first equation

Factor out $x$:
$x(x - 6) = 0$
Set each factor to 0:
$x=0$ or $x-6=0 \implies x=6$

Step3: Rearrange second equation

$0 = 5x^2 - 2x + 6$
Rewrite as standard form:
$5x^2 - 2x + 6 = 0$

Step4: Apply quadratic formula

Quadratic formula: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
Here $a=5$, $b=-2$, $c=6$:
$x=\frac{-(-2)\pm\sqrt{(-2)^2-4(5)(6)}}{2(5)}$
Calculate discriminant:
$\sqrt{4 - 120}=\sqrt{-116}=i\sqrt{4\times29}=2i\sqrt{29}$
Simplify the expression:
$x=\frac{2\pm2i\sqrt{29}}{10}=\frac{1\pm i\sqrt{29}}{5}$

Answer:

First equation: $x=0$ or $x=6$
Second equation: $x=\frac{1\pm i\sqrt{29}}{5}$