High School Mathematics Solutions
- \(f(x) = 4 = 2x^2-3x+5\)
- \( 4 = 2x^2-3x+5\)
- \( 0 = 2x^2-3x+1\)
- By Quadratic Equation we can solve the value of x:
- \( x = \frac{-b+\sqrt{b^2-4ac}}{2a}\)
- \( x_1 = \frac{-(-3)+\sqrt{(-3)^2-4(2)(1)}}{2(2)}\)
Answers:
- \( x_1 = 1\)
- \( x_2 = \frac{1}{2}\)