Algebra 2 (1st Edition)

$$x=\frac{9+\sqrt{241}}{4},\:x=\frac{9-\sqrt{241}}{4}$$
Solving the equation, we find: $$4x^2-18x-40=0\\ x=\frac{-\left(-18\right)+\sqrt{\left(-18\right)^2-4\cdot \:4\left(-40\right)}}{2\cdot \:4}\\ x=\frac{9+\sqrt{241}}{4},\:x=\frac{9-\sqrt{241}}{4}$$