How do you solve $2x^2 +3x + 7 = 0$ ?

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Answer 1 · 2017-10-23T16:59:51

See a solution process below:

We can use the quadratic equation to solve this problem:

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:

$color(red)(2)$ for $color(red)(a)$

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

$color(green)(7)$ for $color(green)(c)$ gives:

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

$x = (-color(blue)(3) +- sqrt(9 - 56))/4$

$x = (-color(blue)(3) +- sqrt(-47))/4$

How do you solve 2x^2 +3x + 7 = 0 2x^2 +3x + 7 = 0 ?