Quadratics with unknowns

Question

If $5x^2 – t = 4x$, and $x$ and $t$ are both positive real numbers. What is $x$ equal to?

How do you find $x$? Is there a specific formula?

Answer 1

If $$ax^2 + bx + c = 0,$$ then you can calculate $x$ to be $$x_{1,2} = \frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$

Answer 2

Hint: Use the quadratic formula:

The solutions to $ax^2+bx+c = 0$ are

$$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$

Answer 3

Write your equation in the form $ax^2 + bx + c = 0.\tag{1}$

Doing so gives you $5x^2-4x – t = 0.\tag{2}$

Now, knowing that to solve for and equation of this form $(1)$, we can use the quadratic equation: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

we have, for your equation $(2)$, $a = 5, \;b = -4,\; c = -t$

$$x = \frac{4 \pm \sqrt{16+20t}}{10}$$

Only one solution results (the $+$ in the $\pm$), since $t>0$ and $x>0.$