How do you solve $2x^2+3x-4=0$ using the quadratic formula?

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Answer 1 · 2016-12-07T01:49:11

$x=(-3+sqrt(41))/4,$ $(-3-sqrt(41))/4$

$2x^2+3x-4=0$ is a quadratic equation in the form $ax^2+bx+c$ , where $a=2$ , $b=3$ , and $c=-4$ .

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

Plug the known values into the equation and solve.

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

$x=(-3+-sqrt(9+32))/4$

$x=(-3+-sqrt(41))/4$

$x=(-3+sqrt(41))/4,$ $(-3-sqrt(41))/4$

How do you solve 2x^2+3x-4=02x^2+3x-4=0 using the quadratic formula?