12x^2 + 20x + 28 solve for x
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solve 12x^2 + 20x + 28 = 0 ?
if so, I would first divide every term by 4
3x^2 + 5x + 7 = 0
Now use the quadratic formula
x = (-5 ± √(25-4(3)(7))/6
= (-5 ± √-59)/6
giving us two complex roots.
To solve the equation 12x^2 + 20x + 28 = 0 for x, we can use the quadratic formula. The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x can be found using the formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our case, a = 12, b = 20, and c = 28. Plugging these values into the quadratic formula, we get:
x = (-(20) ± √((20)^2 - 4(12)(28))) / (2(12))
Simplifying further:
x = (-20 ± √(400 - 1344)) / 24
x = (-20 ± √(-944)) / 24
Since we have a negative value inside the square root, it means that there are no real solutions to the equation. The quadratic equation does not intersect the x-axis, which means there are no values of x that make the equation equal to zero. Therefore, the equation 12x^2 + 20x + 28 = 0 has no real solutions.