Question

solve (2d^{2}+4 = 5d)
options: (\frac{5pm isqrt{7}}{4}), (\frac{-5pmsqrt{57}}{2}), (\frac{5pm isqrt{57}}{4}), (\frac{-5pmsqrt{7}}{4})

Explanation:

Step1: Rearrange the equation

First, we rewrite the given quadratic equation \(2d^{2}+4 = 5d\) in standard form \(ax^{2}+bx + c=0\). Subtract \(5d\) from both sides:
\(2d^{2}-5d + 4=0\)
Here, \(a = 2\), \(b=- 5\), \(c = 4\).

Step2: Apply the quadratic formula

The quadratic formula for a quadratic equation \(ax^{2}+bx + c = 0\) is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
Substitute \(a = 2\), \(b=-5\), \(c = 4\) into the formula:
First, calculate the discriminant \(\Delta=b^{2}-4ac=(-5)^{2}-4\times2\times4=25 - 32=- 7\)
Then, \(d=\frac{-(-5)\pm\sqrt{-7}}{2\times2}=\frac{5\pm i\sqrt{7}}{4}\) (since \(\sqrt{-7}=i\sqrt{7}\))

Answer:

\(\boldsymbol{\frac{5\pm i\sqrt{7}}{4}}\) (corresponding to the first option: \(\frac{5\pm i\sqrt{7}}{4}\))