Question

Hey can you please help me posted picture of question :)

Answer 1

X^2-7x+38=5x+3
x^2-7x+38-5x-3=5x+3-5x-3
x^2-12x+35=0
Factoring:
(x-5)(x-7)=0
Two solutions:
x-5=0→x-5+5=0+5→x=5
x-7=0→x-7+7=0+7→x=7
Answer: x=5 and x=7

Answer: Options B. 7  and D. 5

Answer 2

Answer:
The solutions are: 5 and 7

Explanation:
First we would need to put the equation in the standard form which is:
ax² + bx + c = 0
This can be done as follows:
x² - 7x + 38 = 5x + 3
x² - 7x + 38 - 5x - 3 = 0
x² - 12x + 35 = 0
By comparison:
a = 1
b = -12
c = 35


Now, to get the roots, we would need to use the quadratic formula attached in the images.
By substitution, we would find that:\
either x = $\frac{12+ \sqrt{(-12)^2-4(1)(35)} }{2(1)} = 7$

or x = $\frac{12- \sqrt{(-12)^2-4(1)(35)} }{2(1)} = 5$

Hope this helps :)