Solve the equation by the quadratic formula. Give only real number solutions.
2x2−3x=−6
To solve the equation 2x^2 - 3x = -6 using the quadratic formula, we first need to rewrite the equation in the form ax^2 + bx + c = 0.
The given equation is already in this form, with a = 2, b = -3, and c = -6.
The quadratic formula is given by x = (-b ± √(b^2 - 4ac))/(2a).
Substituting the values of a, b, and c into the formula, we have:
x = (-(-3) ± √((-3)^2 - 4(2)(-6)))/(2(2))
x = (3 ± √(9 + 48))/4
x = (3 ± √57)/4
Therefore, the solutions to the equation 2x^2 - 3x = -6 are x = (3 + √57)/4 and x = (3 - √57)/4.
To solve the equation 2x^2 - 3x = -6 using the quadratic formula, we can write the equation in the form ax^2 + bx + c = 0, where a = 2, b = -3, and c = -6.
The quadratic formula states that the solutions for x can be found using the equation:
x = (-b ± √(b^2 - 4ac)) / (2a)
Substituting the values into the formula, we get:
x = (-(-3) ± √((-3)^2 - 4(2)(-6))) / (2(2))
Simplifying further:
x = (3 ± √(9 + 48)) / 4
x = (3 ± √57) / 4
So, the real number solutions to the equation 2x^2 - 3x = -6 are:
x = (3 + √57) / 4
x = (3 - √57) / 4
To solve the quadratic equation 2x^2 - 3x = -6 using the quadratic formula, we first need to rewrite the equation in the standard form: ax^2 + bx + c = 0.
Given equation: 2x^2 - 3x = -6
By moving the -6 to the left side of the equation, we get:
2x^2 - 3x + 6 = 0
The equation is now in the standard form, with a = 2, b = -3, and c = 6.
Now, let's plug these values into the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Substituting the values from the equation, we get:
x = (-(−3) ± √((-3)^2 - 4(2)(6))) / (2(2))
Simplifying further:
x = (3 ± √(9 - 48)) / 4
x = (3 ± √(-39)) / 4
Since we are looking for only real number solutions, we see that the expression √(-39) is negative under the square root. Therefore, the roots are not real numbers.
Hence, the equation 2x^2 - 3x = -6 does not have any real number solutions.