I'm solving a quadratic equation which is 2x^2-x-10=0 and I know that a=2 b=-1 and c=-10 ans with the problem put it into the formula which is -(-10 +- √ -1^2-4(2)(-10) and when I get my answer I get 79, i know there is no root number for 79. did i do something wrong?

## Quadratic Equation Help

# 5 Answers

^{2}-x-10=0

(2x-5)(x+2)=0

^{2 }- x - 9 = 0 then the discriminant would have been 1 + 72 which is 73 and your roots would have had radicals in them.

It's always a good idea to check your answer by substituting your solution for x back into the original equation to see if it works. This will give you confidence that you have the right answer or let you know if you need to try again. All professionals will do this.

In most practical problems you will rarely get an answer that has a nice whole number as the result. So if you don't get one it still may be correct. Just check it.

Good job on making a solid attempt to solve this problem!

See Nataliya's answer for the full explanation. The key point where you went wrong was on b^{2}. The b value is -1, which means the entire value
*including* the negative sign is being squared, **(-1) ^{2}**=1. So the value under the radical should have been

**1+80=81**and you accidentally ended up with -1+80=79.

ax^{2} + bx + c = 0 ; x_{1,2} = (-b ± √D)/2a ; D = b^{2} - 4ac .

~~~~~~~~~

2x^{2} - x - 10 = 0

D = (-1)^{2} - 4 · 2 · (-10) = 1 + 80 = 81

**x _{1} **= (-(-1) + √81)/4 = (1 + 9)/4 =

**5/2 = 2.5**

**x**(-(-1) - √81)/4 = (1 - 9)/4 =

_{2}=**-2**