 Hi and welcome to the session. I am Arsha and I am going to help you with the following question which says solve the equation for the 7x square minus 10x plus 1 is equal to 0. The general form of a quadratic equation is of deforming x square plus dx plus c is equal to 0 where a is not equal to 0 and the solution of this equation is x is equal to minus b plus minus root over b square minus 4ac gone to a. So, these are some ideas which we are going to use to solve the above problem. So, these are our key ideas. Let us now begin with the solution. The equation is 27, x square minus 10x plus 1 is equal to 0 and I am comparing it with the standard form of the quadratic equation. We find here that a is equal to 27, b is equal to minus 10, c is equal to 1. Now, to find the solution of this quadratic equation, first let us find the value of b square minus 4ac. Now, b square minus 4ac is equal to minus 10 whole square minus 4 into 27 into 1 which is further equal to 100 minus 108 which is equal to minus 8. Now, minus 8 can be written as 8 into minus 1 and minus 1 is outer square. So, b square minus 4ac is 8 outer square. Therefore, root over b square minus 4ac will be root over 8 outer square which is further equal to 2 root 2 outer term. Now, let us find the solution x which is given by minus b plus minus root over b square minus 4ac upon 2a. Now, b is minus 10. So, we have minus of minus 10 plus minus root over b square minus 4ac is root 2 iota upon 2 into a that is 2 into 27 which is further equal to 10 plus minus 2 root 2 iota upon 2 into 27. Now, taking 2 common from the numerator x plus minus root over 2 iota upon 2 into 27. Now, canceling the common multiple we have 5 plus minus root 2 iota upon 27 and now separating the real and imaginary parts we have 5 upon 27 plus minus root over 2 upon 27 iota. Answer to the given equation solving it is 5 upon 27 plus minus root over 2 upon 27 iota. So, this completes the solution. Hope you enjoyed it. Take care and have a good day.