# Math Help - How to solve this equation

1. ## How to solve this equation

I need step by step help on solving this equation:

4x^2 - 15x = 3

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

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

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

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

You got it from here?

3. Originally Posted by pickslides
$\displaystyle 4x^2 - 15x = 3$

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

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

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

You got it from here?
I'm confused on what to do next ...

4. 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.

5. 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!