Question

Solve 16t2 - 96t + 48 = 0.

Answer 1

Answer:

the question is not written correctly 16t2 can't be used in the equation

Step-by-step explanation:

do it it correctly you need to combine the like numbers, subtract the 48 from both sides and then divide the number with the t to both sides.

Answer 2

Answer:

We get the value of t as:

$t=3+\sqrt{6}\ and\ t=3-\sqrt{6}$

Step-by-step explanation:

We are asked to solve the given equation for t.

The quadratic equation in terms of the variable t is given by:

$16t^2-96t+48=0$

On dividing both side of the equation by 16 we get:

$t^2-6t+3=0$

Now, we know that any quadratic equation of the type:

$at^2+bt+c=0$ has solution as :

$t=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

Here,

$a=1,\ b=-6\ and\ c=3$

i.e.

$t=\dfrac{-(-6)\pm \sqrt{(-6)^2-4\times 1\times 3}}{2\times 1}\\\\t=\dfrac{6\pm \sqrt{36-12}}{2}\\\\t=\dfrac{6\pm \sqrt{24}}{2}\\\\t=\dfrac{6\pm 2\sqrt{6}}{2}\\\\t=3\pm \sqrt{6}$

i.e.

$t=3+\sqrt{6}\ and\ t=3-\sqrt{6}$