Question

Use the quadratic formula to solve: \[ 2 x^{2}-5 x-3=0 \]

Answer 1

Answer:
So by using the quadratic formula, we find the answer:
\[
x=3 \quad \text { or } \quad x=-\frac{1}{2}
\]

 

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 \(\quad x=-\frac{1}{2}\)

 

Answer
So by using the quadratic formula, we find the answer:
\[
x=3 \quad \text { or } \quad x=-\frac{1}{2}
\]