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Use the quadratic formula to solve: $2 x^{2}-5 x-3=0$
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$x=3 \quad \text { or } \quad x=-\frac{1}{2}$

Explanation:

Step 1: Find $a, b, c$
Our equation looks like:
$a x^{2}+b x+c=0$
Let's find $a, b, c$ :
- $a$ is the coefficient of $x^{\wedge} 2$
- $b$ is the coefficient of $x$
- $\mathrm{c}$ is the constant term
So for our equation:
$\begin{gathered} 2 x^{2}-5 x-3=0 \\ 2 x^{2}+-5 x+-3=0 \\ a=2 \\ b=-5 \\ c=-3 \end{gathered}$

Step 2: Plug in $a, b, c$
Let's plug in values $a=2, b=-5, c=-3$ to solve $2 x^{\wedge} 2-5 x-3=0$ :
$\begin{gathered} x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\ x=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(2)(-3)}}{2(2)} \end{gathered}$

Step 3: Simplify
We then simplify step-by-step:
$\begin{gathered} x=\frac{5 \pm \sqrt{25-(-24)}}{4} \\ x=\frac{5 \pm \sqrt{49}}{4} \\ x=\frac{5 \pm 7}{4} \end{gathered}$
Separate the plus-minus symbol into two solutions:
$x=\frac{5+7}{4} \quad$ or $\quad x=\frac{5-7}{4}$
$x=\frac{12}{4} \quad$ or $\quad x=\frac{-2}{4}$
$x=3 \quad$ or $\quad x=-\frac{1}{2}$

$x=3 \quad \text { or } \quad x=-\frac{1}{2}$

by Diamond (38,951 points)

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