How do you solve using the quadratic formula $3x^2 + 4x = 6$ ?

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Answer 1 · 2018-03-11T09:06:11

$x=(-4+-2sqrt22)/6$

The quadratic formula says that if we have a quadratic equation in the form:

$ax^2+bx+c=0$

The solutions will be:
$x=(-b+-sqrt(b^2-4ac))/(2a)$

In our case, we have to subtract $6$ from both sides to get it equal to $0$ :

$3x^2+4x-6=0$

Now we can use the quadratic formula:
$x=(-4+-sqrt((-4)^2-4*3*-6))/(2*3)$

$x=(-4+-sqrt(16-(-72)))/6$

$x=(-4+-sqrt(88))/6=(-4+-sqrt(22*4))/6=(-4+-2sqrt22)/6$

How do you solve using the quadratic formula 3x^2 + 4x = 63x^2 + 4x = 6?