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3 votes

2 {x }^(2) - 10x - 4 = 0


User Joma
by
6.7k points

2 Answers

3 votes

Answer:


x_1 = (5 - √(33))/(2)\\x_2 = (5 + √(33))/(2)

That's assuming we were supposed to solve for x.

Explanation:


ax^2 + bx + c = 0\vspace{3pt}\\\Delta = b^2 - 4ac\vspace{4pt}\\x = (-b \pm √(\Delta))/(2a)\\\\2x^2 - 10x - 4 = 0\\x^2 - 5x - 2 = 0\\\Delta = (-5)^2 - 4\cdot 1\cdot (-2) = 25 + 8 = 33\\x_1 = (5 - √(33))/(2)\\x_2 = (5 + √(33))/(2)

User Vadym Stetsiak
by
5.6k points
2 votes

Answer:

see below

Explanation:

I'm assuming you want to find the roots for this equation?

using the quadratic formula:
x_(1,\:2)=(-\left(-10\right)\pm √(\left(-10\right)^2-4\cdot \:2\left(-4\right)))/(2\cdot \:2)


x=(5+√(33))/(2) &
\:x=(5-√(33))/(2)

User Clunven
by
6.1k points