Question

1. $2x^2 - 3x - 15 = 5$

Explanation:

Step1: Rewrite the equation

Subtract 5 from both sides to set the equation to standard quadratic form \(ax^2 + bx + c = 0\).
\(2x^2 - 3x - 15 - 5 = 0\)
\(2x^2 - 3x - 20 = 0\)

Step2: Use the quadratic formula

The quadratic formula is \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b=-3\), and \(c = - 20\).
First, calculate the discriminant \(\Delta=b^2-4ac=(-3)^2-4\times2\times(-20)=9 + 160 = 169\)
Then, find the roots:
\(x=\frac{-(-3)\pm\sqrt{169}}{2\times2}=\frac{3\pm13}{4}\)

Step3: Calculate the two roots

For the plus sign:
\(x_1=\frac{3 + 13}{4}=\frac{16}{4}=4\)
For the minus sign:
\(x_2=\frac{3-13}{4}=\frac{-10}{4}=-\frac{5}{2}=-2.5\)

Answer:

\(x = 4\) or \(x=-\frac{5}{2}\) (or \(x=-2.5\))