Question

How to solve 2x^2-3x-1=0

Answer 1

The given quadratic equation is expressed as

2x^2 - 3x - 1 = 0

We would apply the general formula for quadratic equations which is expressed as

$\begin{gathered} x\text{ = }\frac{-b\text{ +/-}\sqrt{b^2-4ac}}{2a} \\ \text{Looking at the equation,} \\ a\text{ = 2, b = - 3 and c = - 1} \\ \text{If we substitute these values into the formula, it becomes} \\ x\text{ = }\frac{--3\text{ +/-}\sqrt{-3^2-4(2\times-1}}{2\times-1} \\ x\text{ = }\frac{3\text{ +/-}\sqrt{9+8}}{-2} \\ x\text{ = }\frac{3\text{ +/-}\sqrt{17}}{-2} \\ x\text{ = }\frac{3\text{ + }\sqrt{17}}{-\text{ 2}}\text{ or }\frac{3\text{ - }\sqrt{17}}{-\text{ 2}} \\ x\text{ = }\frac{3\text{ + 4.12}}{-\text{ 2}}\text{ or }\frac{3\text{ - 4.12}}{-2} \\ x\text{ = - 3.56 or x = 0.56} \end{gathered}$