How do you solve #2x^2+3x-2=0#?

Question

HIX Tutor · Accepted Answer

undefined To solve the equation 2x^2 + 3x - 2 = 0, follow these steps:

Step 1: Write down the equation: 2x^2 + 3x - 2 = 0
Step 2: Factor the quadratic equation or use the quadratic formula.
Step 3: After factoring or using the quadratic formula, you will find the values of x that satisfy the equation.

Nathan Axel · Answer

#x= 1/2# or #-2# #2x^2+3x-2=0# Factorise #-># #(2x-1)(x+2)=0# #:.# either #(2x-1)=0 -> x=1/2# or #(x+2)=0 -> x=-2#

Eva Adamson · Answer

undefined To solve the quadratic equation $2x^2 + 3x - 2 = 0$, you can use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a$, $b$, and $c$ are the coefficients of the quadratic equation in the form $ax^2 + bx + c = 0$.

For the given equation $2x^2 + 3x - 2 = 0$, the coefficients are $a = 2$, $b = 3$, and $c = -2$. Plugging these values into the quadratic formula, we get:

$$x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2}$$

Simplifying inside the square root gives:

$$x = \frac{-3 \pm \sqrt{9 + 16}}{4}$$
$$x = \frac{-3 \pm \sqrt{25}}{4}$$
$$x = \frac{-3 \pm 5}{4}$$

So, the solutions are:

$$x_1 = \frac{-3 + 5}{4} = \frac{2}{4} = \frac{1}{2}$$

$$x_2 = \frac{-3 - 5}{4} = \frac{-8}{4} = -2$$

Therefore, the solutions to the equation $2x^2 + 3x - 2 = 0$ are $x = \frac{1}{2}$ and $x = -2$.