If the two roots of the quadratic 7x^2+3x+k are \(\frac{-3\pm i\sqrt{299}}{14}\), what is k?
+934 The numbers under the radical are b^2-4ac according to the quadratic formula. "i sqrt299" equals sqrt -299. SO we know b^2-4ac equals -299 and a= 7 and b=3. So the equation we get is 3^2- 4 * 7 * k= -299. From this, we get k=11.
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