The quadratic equation $ax^2+20x+c=0$ has exactly one solution. If $a+c=29$, and $a

a² + c² = 641

Step-by-step explanation:

Given :-

ax² + 20x + c has exactly one solution .a + c = 29 .

For exactly one Solution ,

b² - 4ac = 0 20² - 4*a*c = 0 4ac = 400 ac = 100

Also ,

a + c = 29 ( a + c)² = 29²a² + c² + 2ac = 841 a² + c² + 2*100 = 841a²+ c² = 841 - 200 a² + c² = 641

than do 23.50 times 18 and then divide your answer by 100 than you will get x.

Does the problem say y=x+4?