How do you solve $9x^2 + 6x + 1= 0$ using the quadratic formula?

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Answer 1 · 2017-06-06T21:07:20

See a solution process below:

For $a^x2 + bx + c = 0$ , the values of $x$ which are the solutions to the equation are given by:

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

Substituting $9$ for $a$ ; $6$ for $b$ and $1$ for $c$ gives:

$x = (-6 +- sqrt(6^2 - (4 * 9 * 1)))/(2* 9)$

$x = (-6 +- sqrt(36 - 36))/(2* 9)$

$x = (-6 +- sqrt(0))/(2* 9)$

$x = (-6 +- 0)/(2* 9)$

$x = -6/18$

$x = -1/3$

How do you solve 9x^2 + 6x + 1= 09x^2 + 6x + 1= 0 using the quadratic formula?