Use the quadratic formula to solve for N: 4x^2+x-4=0 Round your answer to the nearest hundredth. if there is more than one solution,separate them with commas x= square

To solve the quadratic equation $4x^{2}+x-4=0$ using the quadratic formula, we can use the formula:<br /><br />$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$<br /><br />where $a$, $b$, and $c$ are the coefficients of the quadratic equation.<br /><br />In this case, $a = 4$, $b = 1$, and $c = -4$.<br /><br />Plugging these values into the formula, we get:<br /><br />$x = \frac{-1 \pm \sqrt{1^2 - 4(4)(-4)}}{2(4)}$<br /><br />Simplifying the expression under the square root:<br /><br />$x = \frac{-1 \pm \sqrt{1 + 64}}{8}$<br /><br />$x = \frac{-1 \pm \sqrt{65}}{8}$<br /><br />Now, we can calculate the two possible solutions for $x$:<br /><br />$x_1 = \frac{-1 + \sqrt{65}}{8} \approx 0.84$<br /><br />$x_2 = \frac{-1 - \sqrt{65}}{8} \approx -1.24$<br /><br />Therefore, the solutions to the quadratic equation $4x^{2}+x-4=0$ are $x = 0.84$ and $x = -1.24$.