2X^minus4Xplus5 how can i solve this?
To solve the equation 2x^2 - 4x + 5, you can use the quadratic formula. The quadratic formula is written as:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this equation, a is the coefficient of x^2 (which is 2), b is the coefficient of x (which is -4), and c is the constant term (which is 5).
Let's substitute these values into the quadratic formula:
x = (-(-4) ± √((-4)^2 - 4(2)(5))) / (2(2))
Simplify the equation further:
x = (4 ± √(16 - 40)) / 4
x = (4 ± √(-24)) / 4
Since the expression inside the square root is negative, it means that the equation has no real solutions. This is because the quadratic equation 2x^2 - 4x + 5 does not intersect the x-axis (no x-intercepts).
Therefore, the equation 2x^2 - 4x + 5 has no solution in the set of real numbers.