How do you solve $x^2-3x+9/4=0$ using the quadratic formula?

Question

5407 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-28T21:48:41

See a solution process below:

For $color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0$ , the values of $x$ which are the solutions to the equation are given by:

$x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))$

Substituting:

$color(red)(1)$ for $color(red)(a)$

$color(blue)(-3)$ for $color(blue)(b)$

$color(green)(9/4)$ for $color(green)(c)$ gives:

$x = (-color(blue)(-3) +- sqrt(color(blue)(-3)^2 - (4 * color(red)(1) * color(green)(9/4))))/(2 * color(red)(1))$

$x = (3 +- sqrt(9 - 9))/2$

$x = (3 +- 0)/2$

$x = 3/2$

How do you solve x^2-3x+9/4=0x^2-3x+9/4=0 using the quadratic formula?