Given the equation - x² + 5x = 3, which can be rewritten as:
- x² + 5x - 3 = 0
where a = -1, b = 5 and c = -3.
Quadratic formula:
![\frac{-b\text{ }\pm\text{ }\sqrt[]{b^2\text{ - 4ac}}}{2a}](https://tex.z-dn.net/?f=%5Cfrac%7B-b%5Ctext%7B%20%7D%5Cpm%5Ctext%7B%20%7D%5Csqrt%5B%5D%7Bb%5E2%5Ctext%7B%20-%204ac%7D%7D%7D%7B2a%7D)
Now, we just replace the values of a, b and c on the equation above.
![\frac{-5\text{ }\pm\text{ }\sqrt[]{5^2\text{ - 4(-1)(3)}}}{2(-1)}](https://tex.z-dn.net/?f=%5Cfrac%7B-5%5Ctext%7B%20%7D%5Cpm%5Ctext%7B%20%7D%5Csqrt%5B%5D%7B5%5E2%5Ctext%7B%20-%204%28-1%29%283%29%7D%7D%7D%7B2%28-1%29%7D)
=
![\frac{5}{2}\text{ }\pm\text{ }\frac{\sqrt[]{13}}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B5%7D%7B2%7D%5Ctext%7B%20%7D%5Cpm%5Ctext%7B%20%7D%5Cfrac%7B%5Csqrt%5B%5D%7B13%7D%7D%7B2%7D)
To find the pre image you need to back track on the image. To get to the image you used (x-6,y+8). Now you need to use the exact opposite to get back to the pre image. For this you would change the signs to look like (x+6,y-8). Now we just apply this to (-4,1).
(-4+6,1-8)
(2,-7) should be the pre image point.
Because the lines are parallel, 75 is equal to 11x-2
75=11x-2
77=11x
x=7
Answer:
10x-5
Step-by-step explanation:
Because u can simplify the equation into a shorter one by simplifying the numbers in paranthesis with 5 and 5 times 2 equal 10 as 5 times 1 equal 5
