 Welcome folks once again to problem solving session on quadratic equations. Now in this video we are going to understand how to solve a quadratic equation by Sridharacharya's rule or which is also called quadratic formula. So let's recap what Sridharacharya's rule said and we will just you know use the formula instead of going through the steps over there. So for how did we arrive at Sridharacharya's formula you can you know watch the previous video and you can understand it from there. So let's solve this quadratic equation. So Sridharacharya formula says that if you have an equation ax square plus bx plus c equals zero where a is not equal to zero and a, b, c are real numbers then the solution is given by x is equal to minus b plus root d by 2a and another another solution is x is equal to minus b minus root d upon 2a where d is equal to b square minus 4ac okay. We will talk about the nature of roots and other things in a later session. So you know without going into that let us first try and see how do we apply this formula. So now it's very clear that in this given equation a is 3, b is 2 root 5 and c is minus 5. So what's the value of d? d will be b square minus 4 times ac is equal to 2 root 5 squared. So it is 2 root 5 squared and minus 4 times a, a is how much? 3 times c, c is minus 5. So hence what will be the value like 2, 4, 5s are 20 and then it is plus 12, 5s are 60 so hence it is 80 is it? So the value of d is 80. So now let us find out the solution. So x is equal to first minus b plus root of b upon twice of a so this will be nothing but minus of b so minus 2 root 5 and then plus root of 80 upon 2a so 2a in 2 is 2 into 3, 6 right. So hence it is nothing but minus 2 root 5 and guys we can you know solve root 80, root 80 is nothing but 16 times 5 right divided by 6 which is nothing but minus 2 root 5 and root 80 can be now written as 4 root 5y, 4 square is 16, 16 times 5, 80. So I am not getting to how will it arrive at this you, you guys would be knowing it by now right. So this is the simplification hence it will be nothing but 2 root 5 by 6 hence nothing but root 5 upon 3. So this would be the first solution. What about the other solution? So simply you know what to change in the formula so minus b now last time we took plus this time will take minus by 2a. So hence minus 2 root 5 and minus root 80 upon 6 which is nothing but minus 2 root 5 minus 4 root 5 divided by 6 which is equal to minus 6 root 5 upon 6 hence minus root 5 so we got this as the solution. So we got two solutions minus root 5 and root 5 by 3 correct you can check it check whether the solutions are correct or not by deploying these values back into the equation for example let's check whether the equation whether we have solved it correctly or not so equation is 3x square plus 2 root 5x minus 5 so let us check here and we had got one solution x is equal x is equal to let's say minus root 5 let's check for this whether x equals to minus root 5 is the solution. So deploy you know instead of x put minus 5 minus root 5 sorry so root 5 square plus 2 times root 5 times minus root 5 and minus 5 so if you calculate this is nothing but minus root 5 square is 5 so 15 and this is nothing but plus 2 times minus 5 root 5 into minus root 5 is 5 minus 5 and minus 5 which is coming out to be 0 hence our solution is correct looks like correct similarly you can deploy the other value and check whether the solution obtained is correct or not so for this the two solutions are x is equal to minus root 5 and x is equal to root 5 upon 3 so two equations you will never get more than two solution for a quadratic equation okay so this is what the solutions are