Question

How do you solve `2x^2 + 6x = 5` by using the quadratic formula?

Answer 1

`x_(1,2) = (-6 +- sqrt(76))/4`

Explanation:

For a general form quadratic equation

`color(blue)(ax^2 + bx + c = 0)`

you can calculate the roots by using the quadratic formula

`color(blue)(x_(1,2) = (_b +- sqrt(b^2 - 4ac))/(2a))`

So, the first thing to do here is rearrange your equation to match the general form. To do that, add `-5` to both sides of the equation

`2x^2 + 6x - 5 = color(red)(cancel(color(black)(5))) - color(red)(cancel(color(black)(5)))`

`2x^2 + 6x - 5 = 0`

In your case, you have `a = 2`, `b = 6`, and `c = -5`. This means that the two solutions will take the form

`x_(1,2) = (-6 +- sqrt(6^2 - 4 * 2 * (-5)))/(2 * 2)`

`x_(1,2) = (-6 +- sqrt(76))/4`

SInce `76` is not a perfect square, the two solutions will be

`x_1 = (-6 - sqrt(76))/4" "` and `" "x_2 = (-6 +- sqrt(76))/4`

You an try out one to see if the calculations are correct

`2 * ((-6-sqrt(76))/4)^2 + 6 * ((-6 - sqrt(76))/4) - 5 = 0`

`2/16 * (36 + 2sqrt(76) + 76) + 3/2 * (-6 - sqrt(76)) -5 = 0`

`14 + color(red)(cancel(color(black)(3/2sqrt(76)))) - 9 - color(red)(cancel(color(black)(3/2sqrt(76)))) - 5 = 0`

`14 - 9 - 5 = 0 color(white)(x) color(green)(sqrt())`