For the given equation;

We shall begin by expanding the parenthesis on the left side, after which we would combine all terms on and move all of them to the left side, which shall yield a quadratic equation. Then we shall solve.
Let us begin by expanding the parenthesis;

Now that we have expanded the left side of the equation, we would have;

We shall now solve the resulting quadratic equation using the quadratic formula as follows;
![\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \text{Where;} \\ a=9,b=-30,c=150 \\ x=\frac{-(-30)\pm\sqrt[]{(-30)^2-4(9)(150)}}{2(9)} \\ x=\frac{30\pm\sqrt[]{900-5400}}{18} \\ x=\frac{30\pm\sqrt[]{-4500}}{18} \\ x=\frac{30\pm\sqrt[]{-900\times5}}{18} \\ x=\frac{30\pm\sqrt[]{-900}\times\sqrt[]{5}}{18} \\ x=\frac{30\pm30i\sqrt[]{5}}{18} \\ \text{Therefore;} \\ x=\frac{30+30i\sqrt[]{5}}{18},x=\frac{30-30i\sqrt[]{5}}{18} \\ \text{Divide all through by 6, and we'll have;} \\ x=\frac{5+5i\sqrt[]{5}}{3},x=\frac{5-5i\sqrt[]{5}}{3} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x%3D%5Cfrac%7B-b%5Cpm%5Csqrt%5B%5D%7Bb%5E2-4ac%7D%7D%7B2a%7D%20%5C%5C%20%5Ctext%7BWhere%3B%7D%20%5C%5C%20a%3D9%2Cb%3D-30%2Cc%3D150%20%5C%5C%20x%3D%5Cfrac%7B-%28-30%29%5Cpm%5Csqrt%5B%5D%7B%28-30%29%5E2-4%289%29%28150%29%7D%7D%7B2%289%29%7D%20%5C%5C%20x%3D%5Cfrac%7B30%5Cpm%5Csqrt%5B%5D%7B900-5400%7D%7D%7B18%7D%20%5C%5C%20x%3D%5Cfrac%7B30%5Cpm%5Csqrt%5B%5D%7B-4500%7D%7D%7B18%7D%20%5C%5C%20x%3D%5Cfrac%7B30%5Cpm%5Csqrt%5B%5D%7B-900%5Ctimes5%7D%7D%7B18%7D%20%5C%5C%20x%3D%5Cfrac%7B30%5Cpm%5Csqrt%5B%5D%7B-900%7D%5Ctimes%5Csqrt%5B%5D%7B5%7D%7D%7B18%7D%20%5C%5C%20x%3D%5Cfrac%7B30%5Cpm30i%5Csqrt%5B%5D%7B5%7D%7D%7B18%7D%20%5C%5C%20%5Ctext%7BTherefore%3B%7D%20%5C%5C%20x%3D%5Cfrac%7B30%2B30i%5Csqrt%5B%5D%7B5%7D%7D%7B18%7D%2Cx%3D%5Cfrac%7B30-30i%5Csqrt%5B%5D%7B5%7D%7D%7B18%7D%20%5C%5C%20%5Ctext%7BDivide%20all%20through%20by%206%2C%20and%20we%27ll%20have%3B%7D%20%5C%5C%20x%3D%5Cfrac%7B5%2B5i%5Csqrt%5B%5D%7B5%7D%7D%7B3%7D%2Cx%3D%5Cfrac%7B5-5i%5Csqrt%5B%5D%7B5%7D%7D%7B3%7D%20%5Cend%7Bgathered%7D)
ANSWER:
The answer is 1738
Hope this helps you
Answer:
I think grater then?
Step-by-step explanation:
Answer:
3rd Option: 1
Step-by-step explanation:
Step 1: Define
(x + 4)/3x
x = 2
Step 2: Substitute and Evaluate
(2 + 4)/3(2)
6/6
1
Answer: Infinite
Step-by-step explanation:
We know that in a triangle the sum of all the interior angles must be 180°.
The given angles 50º, 90º and 40º
The sum of the angles 50º+ 90º + 40º= 180°
Thus, a triangle is possible with the given measurement.,
Let there is another triangle with the given angles, then by AAA similarity criteria they are similar.
Similarly, all the triangles with the same measurements of the angles must be similar.
Therefore, there are infinite number of triangles can be possible with angles measuring 50º, 90º, and 40º.