Question

How do you solve `6x^2-8x-3=0` using the quadratic formula?

Answer 1

`x_(1,2) = (-color(blue)8+- sqrt(color(blue)(64)-72))/(12)= -2/3 +- isqrt(2)/6 = 1/3(-2+-sqrt(2)/2)`
Note this means:
`x_1 = 1/3 (-2+sqrt(2)/2)` and `x_1 = 1/3 (-2-sqrt(2)/2)`

Explanation:

I suggest you commit to memory the "Quadratic Formula" it probably one of the formula that you absolutely positively must know by heart: So here is your Quadratic Formula:
Given a 2nd Order Polynomial,`P_2`:

`P_2 = color(red)ax^2 + color(blue)bx + color(green)c` the Roots or solutions to the equation
`x_1` and `x_2` are given by the Quadratic Formula:

`x_(1,2) = (-color(blue)b+- sqrt(color(blue)(b^2)-4color(red)acolor(green)c))/(2color(red)a)`

Now for your equation: `color(red)6x^2-color(blue)8x-color(green)3=0`

`x_(1,2) = (-color(blue)8+- sqrt(color(blue)(8^2)-4*color(red)6*color(green)3))/(2*color(red)6)`
`x_(1,2) = (-color(blue)8+- sqrt(color(blue)(64)-72))/(12)= -2/3 +- isqrt(2)/6 = 1/3(-2+-sqrt(2)/2)`

Answer 2

`x=(4+-sqrt(34))/6`

Explanation:

`1`. Since the given equation is already in standard form, identify the `color(blue)a,color(darkorange)b,` and `color(violet)c` values. Then plug the values into the quadratic formula to solve for the roots.

`color(blue)6x^2` `color(darkorange)(-8)x` `color(violet)(-3)=0`

`color(blue)(a=6)color(white)(XXXXX)color(darkorange)(b=-8)color(white)(XXXXX)color(violet)(c=-3)`

`x=(-b+-sqrt(b^2-4ac))/(2a)`

`x=(-(color(darkorange)(-8))+-sqrt((color(darkorange)(-8))^2-4(color(blue)6)(color(violet)(-3))))/(2(color(blue)6))`

`x=(8+-sqrt(64+72))/12`

`x=(8+-sqrt(136))/12`

`x=(8+-2sqrt(34))/12`

`2`. Factor out `2` from the numerator and denominator.

`x=(2(4+-sqrt(34)))/(2(6))`

`x=(color(red)cancelcolor(black)2(4+-sqrt(34)))/(color(red)cancelcolor(black)2(6))`

`color(green)(|bar(ul(color(white)(a/a)x=(4+-sqrt(34))/6color(white)(a/a)|)))`