T. Pitagora twice => new street = 2

= 8.94 miles;
135*8.94 = 1206.9$;
Combine like terms... 2s and 4s are like terms so that's 6s 4,2,3 are like terms add that up its 9. So it would look like this 6s+9
To get rid of

, you have to take the third root of both sides:
![\sqrt[3]{x^{3}} = \sqrt[3]{1}](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7Bx%5E%7B3%7D%7D%20%3D%20%5Csqrt%5B3%5D%7B1%7D%20)
But that won't help you with understanding the problem. It is better to write

as a product of 2 polynomials:

From this we know, that

is the solution. Another solutions (complex roots) are the roots of quadratic equation.
Answer:
a. x = 2 , x = -5
b. no real roots
Step-by-step explanation:
<u>a. x² + 3x – 10</u>
= x² - 2x + 5x - 10
= x (x²/x - 2x/x) + 5 (5x/5 - 2*5/5)
= (x - 2) (x + 5)
x - 2 = 0 x + 5 = 0
+ 2 +2 -5 -5
-------------- ---------------
x = 2 x = -5
x = 2 , x = -5
<u>b. 2x² - 4x + 3 </u>
This can't be solved because the equation has no real roots, and also the discriminant is negative.