Now, we can identify the values of a, b, and c.
2x^2+3x-2=0 ⇕ 2x^2+ 3x+( - 2)=0
We see that a= 2, b= 3, and c= - 2. Let's substitute these values into the Quadratic Formula.
The solutions for this equation are x= - 3± 54. Let's separate them into the positive and negative cases.
x=- 3± 5/4
x_1=- 3+ 5/4
x_2=- 3- 5/4
x_1=2/4
x_2=- 8/4
x_1=1/2
x_2=- 2
Using the Quadratic Formula, we found that the solutions of the given equation are x_1= 12 and x_2=- 2.
b We want to find the real solutions of the given quadratic equation. Let's begin by isolating the square on one side. Here, this means subtracting 4 from both sides of the equation.
(2x-3)^2+4=0 ⇔ (2x-3)^2= - 4
Notice that there is no real number that results in a negative number when squared. Since - 4 <0, there is no real number that can satisfy the given quadratic equation. Therefore, this equation has no real solutions.
