(-5 plus or minus the square root of 11) divided by 2
x = -0.8416876048223001
x = -4.1583123951777
A quadratic function is ax2+bx+c You can solve for x by using the quadratic formula, which, as the formula requires the use of square roots, would be tricky to put here.
If: x2+x = 12 Then: x2+x-12 = 0 And using the quadratic formula: x = -4 or x = 3
Using the quadratic formula, you get the complex answers of 1 + 2i and 1 - 2i
To find the solution to this equation, you need to rearrange the terms and solve for the variable. 4 = 2b + b^2 can be rewritten as b^2 + 2b - 4 = 0. You can then solve this quadratic equation by factoring, completing the square, or using the quadratic formula.
X= 1.567764363, -9.567764363
Using the quadratic formula, I found the solution set is x=2,x=-9
Using the quadratic equation formula: x = -5-/+ the square root of 7
It is a quadratic equation and its solutions can be found by using the quadratic equation formula.
Without an equality sign the given expression can't be considered to be an equation but if it equals 0 then using the quadratic equation formula will give its solutions.
A quadratic function is ax2+bx+c You can solve for x by using the quadratic formula, which, as the formula requires the use of square roots, would be tricky to put here.
If: x2+x = 12 Then: x2+x-12 = 0 And using the quadratic formula: x = -4 or x = 3
Using the quadratic formula, you get the complex answers of 1 + 2i and 1 - 2i
The equation does not have a real number solution. Using the quadratic formula will give it's conjugate pair complex solution.
To find the solution to this equation, you need to rearrange the terms and solve for the variable. 4 = 2b + b^2 can be rewritten as b^2 + 2b - 4 = 0. You can then solve this quadratic equation by factoring, completing the square, or using the quadratic formula.
X= 1.567764363, -9.567764363
Yes
The solution to a math problem involving a quadratic equation is the values of the variable that make the equation true, typically found using the quadratic formula or factoring.