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Use the quadratic formula. In x2 - 4x + 29 = 0, a is 1, b is -4, c = 29. The quadratic formula is: x = (-b plusminus squareroot(b2 - 4ac)) / 2a

Use the quadratic formula. In x2 - 4x + 29 = 0, a is 1, b is -4, c = 29. The quadratic formula is: x = (-b plusminus squareroot(b2 - 4ac)) / 2a

Use the quadratic formula. In x2 - 4x + 29 = 0, a is 1, b is -4, c = 29. The quadratic formula is: x = (-b plusminus squareroot(b2 - 4ac)) / 2a

Use the quadratic formula. In x2 - 4x + 29 = 0, a is 1, b is -4, c = 29. The quadratic formula is: x = (-b plusminus squareroot(b2 - 4ac)) / 2a

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Use the quadratic formula. In x2 - 4x + 29 = 0, a is 1, b is -4, c = 29. The quadratic formula is: x = (-b plusminus squareroot(b2 - 4ac)) / 2a

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Q: Find all complex solutions for the equation x2 equals 4x - 29?
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How can you find the complex solutions of any quadratic equation?

I suggest you use the quadratic formula.


How do you factor x2-4x-80 equals 0?

Use the quadratic equation formula to find the solutions to this equation.


What statement must be true of an equation before you can use the quadratic formula to find the solutions?

The quadratic formula can be used to find the solutions of a quadratic equation - not a linear or cubic, or non-polynomial equation. The quadratic formula will always provide the solutions to a quadratic equation - whether the solutions are rational, real or complex numbers.


How many solutions are there in the equation 2x equals 10?

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What is the answer for x x -6x-8 equals -56?

x² - 6x -8 = -56 <=> x² -6x + 48 = 0 We now calculate the discriminant (which equals b²-4ac for an equation of the form ax² + bx + c-: D = b²-4ac = (-6)² - 4*1*48 = 36 - 192 = -156 I don't know what kind of course in calculus you are taking, but if you only want the real answers to this equation, you can stop here because the discriminant is negative, meaning there are no real solutions. However, there are complex solutions to this equation The complex roots of D are sqrt(156)*i=12,5*i and -sqrt(156)*i=-12,5*i There are two solutions to a quadratic equation, namely: x1 = (-b + sqrt(D))/(2*a) and x2 = (-b - sqrt(D))/(2*a) so the two solutions we find for this equation are: x1 = (-b + sqrt(D))/(2*a) = (6+12,5i)/2 = 3+6,25i and x2 = (-b - sqrt(D))/(2*a) = (6-12,5i)/2 = 3-6,25i x1 and x2 are complex solutions to this quadratic equation.


Find three different ordered pairs that are solutions of the equation y equals 2x-1?

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How do you find the number of solutions in a quadratic equation?

Factoring by the AC method, difference of squares, perfect square trinomial. If not factorable by those ways, you can use the quadratic formula. You can also find zeros by synthetic division. If there are not any real solutions, then the solutions are said to be complex, they do not cross the x axis.


How do you find quadratic equation with only 2 solutions?

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Can you solve x plus 2y equals 2?

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What statements must be true of an equation before you can use the quadratic formula to find the solutions?

That the discriminant of the quadratic equation must be greater or equal to zero for it to have solutions. If the discriminant is less than zero then the quadratic equation will have no solutions.


How you find the solution of a quadratic equation by graphing its quadratic equation?

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