 Hi and welcome to the session. I am Arshad and I am going to help you with the following question which says solve the equation 21x square minus 28x plus 10 is equal to 0. Now a quadratic equation is of the form ax square plus bx plus c is equal to 0 where a is not equal to 0 and the solution of this equation is given by x is equal to minus b plus minus u to the power b square minus 4ac upon 2a. So these are some ideas which we are going to use to solve the above problem. So this is our key idea. Let us now begin with the solution and the given equation is 21x square minus 28x plus 10 is equal to 0. On comparing it with the general form of the quadratic equation we find here a is equal to 21, b is equal to minus 28 and c is equal to 10. Let us first find out the value of b square minus 4ac to find the solution. b square minus 4ac is equal to 28, the negative sign 4 square minus 4 into 8 as 21 into c just 10 which is further equal to 28 into 28 is 784 minus into minus is plus and minus 840 is equal to minus 56. Now minus 56 can be written as 56 into minus 1 and minus 1 is iota square so 56 iota square and thus root over b square minus 4ac is equal to root over 56 iota square which can be written as 2 root over 14 iota. Now let us find the solutions which are given by x is equal to minus b plus minus root over b square minus 4ac upon 2a. Now b is minus 28 plus minus root over b square minus 4ac is 2 root over 14 iota upon 2 into 21 which is equal to 28 plus minus 2 root over 14 iota upon 2 into 21 and now taking 2 common from the numerator we have 2 into 14 plus minus root over 14 iota upon 2 into 21 and now canceling the common multiples we have 14 plus minus root over 14 iota upon 21 and now separating the real and imaginary parts we have 14 upon 21 plus minus root over 14 upon 21 iota and thus the solution of the given equation is 14 upon 21 plus minus root over 14 upon 21 iota. So this completes the solution hope you enjoyed it take care and have a good day.