How to solve this equation

**jenbones** · September 1st 2010, 04:34 PM

I need step by step help on solving this equation:

4x^2 - 15x = 3

**pickslides** · September 1st 2010, 04:41 PM

$\displaystyle 4x^2 - 15x = 3$

$\displaystyle 4x^2 - 15x -3= 0$

by the quadartic formula

$\displaystyle ax^2 + bx +c= 0 \implies x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$\displaystyle 4x^2 - 15x -3= 0 \implies x = \frac{-(-15)\pm\sqrt{(-15)^2-4\times 4\times (-3)}}{2\times 4}$

You got it from here?

**jenbones** · September 1st 2010, 05:10 PM

Originally Posted by pickslides

$\displaystyle 4x^2 - 15x = 3$

$\displaystyle 4x^2 - 15x -3= 0$

by the quadartic formula

$\displaystyle ax^2 + bx +c= 0 \implies x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$\displaystyle 4x^2 - 15x -3= 0 \implies x = \frac{-(-15)\pm\sqrt{(-15)^2-4\times 4\times (-3)}}{2\times 4}$

You got it from here?

I'm confused on what to do next ...

**fireballs619** · September 1st 2010, 05:24 PM

given the quadratic formula with all variables filled in, you simply simplify down (notice the pun there)

$x = \frac{-(-15)\pm\sqrt{(-15)^2-4\times 4\times (-3)}}{2\times 4}$

$x = \frac{15\pm\sqrt{225-48}}{8}$

$x = \frac{15\pm\sqrt{177}}{8}$

Using order of operations, of course.

**jenbones** · September 1st 2010, 07:40 PM

Originally Posted by fireballs619

given the quadratic formula with all variables filled in, you simply simplify down (notice the pun there)

$x = \frac{-(-15)\pm\sqrt{(-15)^2-4\times 4\times (-3)}}{2\times 4}$

$x = \frac{15\pm\sqrt{225-48}}{8}$

$x = \frac{15\pm\sqrt{177}}{8}$

Using order of operations, of course.

thanks!

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