Question

Use the quadratic formula to solve. Round to the nearest tenth if necessary. 2x^2 -3x -7 = 0

Answer 1

The quadratic formula is:

$\frac{-b + \sqrt{b^2 - 4ac} }{2a}$ and  $\frac{-b - \sqrt{b^2 - 4ac} }{2a}$

In this case, a = 2, b = -3, and c = -7

So, we can plug in the numbers to get:

  $\frac{-(-3) + \sqrt{(-3)^2 - 4(2)(-7)} }{2(2)}$ and    $\frac{-(-3) - \sqrt{(-3)^2 - 4(2)(-7)} }{2(2)}$

Simplifying, we get:

    $\frac{3 + \sqrt{65} }{4}$ and   $\frac{3 - \sqrt{65} }{4}$

You need to use a calculator to find what these would be in decimal form. The answer, rounded to the nearest tenth, is: x = 2.8, x = -1.3