Question

The solution of a quadratic equation can be found by graphing each side separately and locating the points of intersection. You may wish to consult page 532 for help in approximating solutions. $$ -x^{2}-2=4 x^{2}+6 x-3 $$

Short answer

The solutions of the equation are \(x1 = \frac{-6 + \sqrt{56}}{10}\) and \(x2 = \frac{-6 - \sqrt{56}}{10}\).

Step-by-step solution

Bring All Terms to One Side of Equation

The equation can be brought into the standard form by bringing all terms to one side. So, -x^2 - 2 = 4x^2 + 6x - 3 becomes 0 = 5x^2 + 6x -1.

Apply Quadratic Formula

The quadratic formula is defined as \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\) . For the quadratic equation \(5x^2+6x-1=0\), a = 5, b = 6, and c = -1. Substituting these values into the quadratic formula yields the solutions.

Calculate Values of x

Using the quadratic formula, we get \(x = \frac{-6 \pm \sqrt{6^2 - 4*5*-1}}{2*5}\), which simplifies to \(x = \frac{-6 \pm \sqrt{36 + 20}}{10}\). Further simplification gives \(x = \frac{-6 \pm \sqrt{56}}{10}\). Therefore, we get \(x1 = \frac{-6 + \sqrt{56}}{10}\) and \(x2 = \frac{-6 - \sqrt{56}}{10}\).