Question

solve the quadratic equation.
\\( 2x^2 + 9 = 6x \\)
write one exact, simplified solution in each box. you can add or remove boxes. if a solution is not a real number, write it in the form \\( a + bi \\) or \\( a - bi \\), where \\( a \\) and \\( b \\) are real numbers.

Explanation:

Step1: Rewrite in standard form

First, we rewrite the quadratic equation \(2x^{2}+9 = 6x\) in standard form \(ax^{2}+bx + c=0\). Subtract \(6x\) from both sides:
\(2x^{2}-6x + 9=0\)
Here, \(a = 2\), \(b=- 6\), \(c = 9\).

Step2: Use quadratic formula

The quadratic formula is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). Substitute the values of \(a\), \(b\), and \(c\) into the formula:
First, calculate the discriminant \(\Delta=b^{2}-4ac=(-6)^{2}-4\times2\times9=36 - 72=- 36\)
Then, \(x=\frac{-(-6)\pm\sqrt{-36}}{2\times2}=\frac{6\pm6i}{4}=\frac{3\pm3i}{2}\)

Answer:

\(\frac{3 + 3i}{2}\), \(\frac{3 - 3i}{2}\)