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

Question

23267 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-07-15T12:49:10

$x=(-3+sqrt 33)/4,$ $(-3-sqrt 33)/4$

$2x^2+3x-3=0$ is a quadratic equation in the form of $ax^2+bx+c$ , where $a=2,$ $b=3,$ and $c=-3$ .

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

Substitute the given values into the formula.

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

$x=(-3+-sqrt(9+24))/4$ =

$x=(-3+-sqrt33)/4$

Solve for $x$ .

$x=(-3+sqrt33)/4$

$x=(-3-sqrt33)/4$

How do you solve 2x^2+3x-3=02x^2+3x-3=0 using the quadratic formula?