x'2+ 10= 3x+17. Use the quadratic formula to solve.
x'2+ 10= 3x+17
it is generally accepted to use ^ to show exponents.
x^2+ 10= 3x+17
x2 - 3x - 7 = 0
x = (3 ± √(9-4(1)(-7))/2
can you finish it ?
the square root is impossible to do how do you solve it?
To solve the equation x^2 + 10 = 3x + 17 using the quadratic formula, we need to rewrite the equation in the standard quadratic form ax^2 + bx + c = 0, where a, b, and c are coefficients.
Here's how to use the quadratic formula:
1. Move all terms to one side to set the equation to zero:
x^2 - 3x + 7 = 0
2. Identify the values of a, b, and c:
a = 1 (coefficient of x^2)
b = -3 (coefficient of x)
c = 7 (constant term)
3. Substitute the values of a, b, and c into the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
Plugging in the values:
x = (-(-3) ± √((-3)^2 - 4(1)(7))) / (2 * 1)
4. Simplify the equation inside the square root:
x = (3 ± √(9 - 28)) / 2
x = (3 ± √(-19)) / 2
Since the expression inside the square root is negative (-19), the quadratic does not have real solutions. This means that the equation x^2 + 10 = 3x + 17 has no real solutions.