This is a quadratic equation, i.e. an equation involving a polynomial of degree 2. To solve them, you must rearrange them first, so that all terms are on the same side, so we get
![x^2 + 11x - 4 = 0](https://tex.z-dn.net/?f=%20x%5E2%20%2B%2011x%20-%204%20%3D%200%20)
i.e. now we're looking for the roots of the polynomial. To find them, we can use the following formula:
![x_{1,2} = \frac{-b\pm\sqrt{b^2-4ac}}{2}](https://tex.z-dn.net/?f=%20x_%7B1%2C2%7D%20%3D%20%5Cfrac%7B-b%5Cpm%5Csqrt%7Bb%5E2-4ac%7D%7D%7B2%7D%20)
where
is a compact way to indicate both solutions
and
, while
are the coefficients of the quadratic equation, i.e. we consider the polynomial
.
So, in your case, we have ![a=1,\ \ b=11,\ \ c=-4](https://tex.z-dn.net/?f=%20a%3D1%2C%5C%20%5C%20b%3D11%2C%5C%20%5C%20c%3D-4%20)
Plug those values into the formula to get
![x_{1,2} = \frac{-11\pm\sqrt{121+16}}{2} = \frac{-11\pm\sqrt{137}}{2}](https://tex.z-dn.net/?f=%20x_%7B1%2C2%7D%20%3D%20%5Cfrac%7B-11%5Cpm%5Csqrt%7B121%2B16%7D%7D%7B2%7D%20%3D%20%5Cfrac%7B-11%5Cpm%5Csqrt%7B137%7D%7D%7B2%7D%20)
So, the two solutions are
![x_1 = \frac{-11+\sqrt{137}}{2}](https://tex.z-dn.net/?f=%20x_1%20%3D%20%5Cfrac%7B-11%2B%5Csqrt%7B137%7D%7D%7B2%7D%20)
![x_2 = \frac{-11-\sqrt{137}}{2}](https://tex.z-dn.net/?f=%20x_2%20%3D%20%5Cfrac%7B-11-%5Csqrt%7B137%7D%7D%7B2%7D%20)
Answer:
Solve y=x^2+2 for x.A.B.C.D
Step-by-step explanation:
sorry lamo
4/5 is the answer.
The formula is y=mx+b, where m is the slope
Answer:
2.3 could, but not -0.25.
Step-by-step explanation:
It is impossible to have any probability larger than 100 or lower than 0. 2.3 is possible bc it's below 100 and above 0.
It would be what is 20% of 70. x = 20/100 x 70 = 14
20% of 70 is 14