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

User Manik Sidana
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1 Answer

18 votes
18 votes

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{ +/-}√(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{ +/-}√(-3^2-4(2*-1)}{2*-1} \\ x\text{ = }\frac{3\text{ +/-}√(9+8)}{-2} \\ x\text{ = }\frac{3\text{ +/-}√(17)}{-2} \\ x\text{ = }\frac{3\text{ + }√(17)}{-\text{ 2}}\text{ or }\frac{3\text{ - }√(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}

User Naufraghi
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