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Q: 9       \small 21x^2-28x+10=0

 

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Given equation is
\small 21x^2-28x+10=0 
Now, we know that the roots of the quadratic equation are given by the formula
\frac{-b\pm \sqrt{b^2-4ac}}{2a}
In this case the value of   a=21,b=-28 \ and \ c= 10
Therefore,
\frac{-(-28)\pm \sqrt{(-28)^2-4.21.10}}{2.21}= \frac{28\pm\sqrt{784-840}}{42} = \frac{28\pm\sqrt{-56}}{42}=\frac{28\pm i2\sqrt{14}}{42}=\frac{2}{3}\pm i\frac{\sqrt{14}}{21}
Therefore, the solutions of requires equation are 

  \frac{2}{3}\pm i\frac{\sqrt{14}}{21}

Posted by

Gautam harsolia

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