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CCSS.Math: ,

we're asked to solve the quadratic equation negative 3x squared plus 10x minus 3 is equal to 0 and it's already written in standard form and there's many ways to solve this but in particular I'll solve it using the quadratic formula so let me just rewrite it we have negative 3x squared plus 10x minus 3 is equal to 0 and actually solve it twice using the quadratic formula to show you that as long as we manipulated this in the valid way the quadratic formula will give us the exact same roots or the exact same solutions to this equation so in this form right over here what are our ABCs well let's just remind ourselves what the quadratic formula even is actually that's a good place to start the quadratic formula tells us that if we have a quadratic equation in the form ax squared plus BX plus C is equal to 0 so in standard form then the roots of this our X are equal to negative B plus or minus the square root of b squared minus 4ac all of that over 2a all of that over 2a and this is derived from completing the square in a general way so it's no magic here and I've derived it in other videos but this is the kwatak form and this is actually giving you two solutions because you have the plot the positive square root here and the negative square root so let's apply it here in the case where in this case a is equal to negative 3 a is equal to negative 3 B is equal to 10 B is equal to 10 and C is equal to negative 3 C is equal to negative 3 so applying the quadratic formula right here we get our solutions to be X is equal to negative B B is 10 so negative B is 10 negative 10 plus or minus the square root of B squared B is 10 so B squared is 100 minus 4 times a times C so minus 4 times negative 3 times negative 3 let me just write it down minus 4 times negative 3 times negative 3 all of that's under the radical sign and then all of that is over 2a so 2 times a is negative 6 so this is going to be equal to negative 10 plus or minus the square root of 100 minus the retime negative 3 times negative 3 is positive 9 positive 9 times 4 is positive 36 we have a minus sign out here so minus 36 all of that over negative 6 this is equal to 100 minus 36 is 64 so negative 10 plus or minus the square root of 64 all of that over negative 6 the principal square root of 64 is 8 we're taking the positive and negative square roots so this is negative 10 plus or minus 8 over negative 6 so if we take the positive version we say X could be equal to negative 10 plus 8 is negative 2 over negative 6 so that was taking the plus version that's this right over here and negative 2 over negative 6 is equal to 1/3 if we take the negative square root negative 10 minus 8 so let's take negative 10 minus 8 that would be X is equal to negative 10 minus 8 is negative 18 and that's going to be over negative 6 over negative 6 negative 18 divided by negative 6 is positive 3 so the two roots for this quadratic equation are positive 1/3 and positive 3 and I want to show you that we'll get the same answer even if we manipulate this some people might not like the fact that our first coefficient here is a negative 3 maybe they want a positive 3 so to get rid of that negative 3 they can multiply both sides of this equation times negative 1 and then if you did that you would get 3x squared minus 10x plus 3 is equal to 0 times negative 1 which is still equal to 0 so in this case a is equal to 3 B is equal to negative 10 and C is equal to 3 again and we can apply the quadratic formula we get X is equal to negative B B is negative 10 so negative negative 10 is positive 10 plus or minus the square root of B squared which is negative 10 squared which is 100 minus 4 times a times C a times C is 9 times 4 is 36 so minus 36 all of that over 2 times a all of that over 6 so this is equal to 10 plus or minus the square root of 64 or really that's just going to be 8 all of that over six if we add 8 here we get 10 plus 8 is 18 over 6 we get X could be equal to 3 or if we take the if we take the negative square root of the negative 8 here 10 minus 8 is 2 2 over 6 is 1/3 so once again you get the exact same solutions