How do you solve $6x^2−7x−3=0$ using the quadratic formula?

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Answer 1 · 2018-05-24T12:11:30

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:

$color(red)(6)$ for $color(red)(a)$

$color(blue)(-7)$ for $color(blue)(b)$

$color(green)(-3)$ for $color(green)(c)$ gives:

$x = (-color(blue)(-7) +- sqrt(color(blue)(-7)^2 - (4 * color(red)(6) * color(green)(-3))))/(2 * color(red)(6))$

$x = (7 +- sqrt(49 - (-72)))/12$

$x = (7 +- sqrt(49 + 72))/12$

$x = (7 +- sqrt(121))/12$

$x = (7 +- 11)/12$

$x = (7 - 11)/12$ ; $x = (7 + 11)/12$

$x = (-4)/12$ ; $x = 18/12$

$x = -1/3$ ; $x = 3/2$

How do you solve 6x^2−7x−3=0 6x^2−7x−3=0 using the quadratic formula?