How do you solve $3x^2+9x-12=0$ using the quadratic formula?

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Answer 1 · 2016-02-15T06:11:26

Solution is $x=1 or -4$ .

Solution of quadratic equation $ax^2+bx+c=0$ is given by

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

Here $a=3, b=9 and c=-12$ . Putting these values

$x=(-9+-sqrt(9^2-4*3*(-12)))/(2*3)$

= $x=(-9+-sqrt(81+144))/6$

= $x=(-9+-sqrt(225))/6$ = $x=(-9+-15)/6$

= $1 or -4$

How do you solve 3x^2+9x-12=03x^2+9x-12=0 using the quadratic formula?