\( 2 x^{2}=-x+5( \) Are the roots real or Unreal

Solve the quadratic equation by following steps: - step0: Solve using the quadratic formula: \(2x^{2}+x-5=0\) - step1: Solve using the quadratic formula: \(x=\frac{-1\pm \sqrt{1^{2}-4\times 2\left(-5\right)}}{2\times 2}\) - step2: Simplify the expression: \(x=\frac{-1\pm \sqrt{1^{2}-4\times 2\left(-5\right)}}{4}\) - step3: Simplify the expression: \(x=\frac{-1\pm \sqrt{41}}{4}\) - step4: Separate into possible cases: \(\begin{align}&x=\frac{-1+\sqrt{41}}{4}\\&x=\frac{-1-\sqrt{41}}{4}\end{align}\) - step5: Rewrite the fraction: \(\begin{align}&x=\frac{-1+\sqrt{41}}{4}\\&x=-\frac{1+\sqrt{41}}{4}\end{align}\) - step6: Rewrite: \(x_{1}=-\frac{1+\sqrt{41}}{4},x_{2}=\frac{-1+\sqrt{41}}{4}\) To determine whether the roots of the equation \( 2x^{2} = -x + 5 \) are real or unreal, we first rearranged the equation into standard form: \[ 2x^{2} + x - 5 = 0 \] Next, we solved this quadratic equation and found the roots: \[ x_{1} = -\frac{1 + \sqrt{41}}{4}, \quad x_{2} = \frac{-1 + \sqrt{41}}{4} \] Since the roots involve the square root of a positive number (\(\sqrt{41}\)), both roots are real numbers. Therefore, the roots of the equation are real.