Question

NEED ANSWER ASAP - 100 POINTS

Answer 1

Using the quadratic formula

The quadratic formula can help us solve for x in a quadratic equation:

$x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$

We can use this when the given equation is organized in standard form: $ax^2+bx+c=0$

Applying this to the problem

$2x^2+5x-4=0$

First we can identify the values of a, b and c:

a = 2

b = 5

c = -4

Plug these values into the quadratic formula:

$x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$

$x=\dfrac{-5 \pm \sqrt{5^2-4(2)(-4)}}{2(2)}$

$x=\dfrac{-5 \pm \sqrt{25+32}}{4}$

$x=\dfrac{-5 \pm \sqrt{57}}{4}$

Answer

Therefore, the two possible values of x are:

$x=\dfrac{-5 +\sqrt{57}}{4}$

$x=\dfrac{-5 -\sqrt{57}}{4}$