# How do you find the roots, real and imaginary, of #y=(x/5-5)(-3x-2)# using the quadratic formula?

See a solution process below:

First, we need to write this equation in standard form:

We can now use the quadratic equation to solve the equation. The quadratic formula states:

Substituting:

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To find the roots, real and imaginary, of ( y = \left(\frac{x}{5} - 5\right)(-3x - 2) ) using the quadratic formula, we first expand and simplify the equation to get it in standard quadratic form, ( ax^2 + bx + c = 0 ).

Expanding and simplifying the given equation:

[ y = \left(\frac{x}{5} - 5\right)(-3x - 2) ] [ y = -\frac{3x^2}{5} + 15x + 10 ]

Now, we have ( a = -\frac{3}{5} ), ( b = 15 ), and ( c = 10 ).

Using the quadratic formula:

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

Substitute the values of ( a ), ( b ), and ( c ) into the formula:

[ x = \frac{-15 \pm \sqrt{15^2 - 4(-\frac{3}{5})(10)}}{2(-\frac{3}{5})} ]

Simplify and solve:

[ x = \frac{-15 \pm \sqrt{225 + 24}}{-\frac{6}{5}} ] [ x = \frac{-15 \pm \sqrt{249}}{-\frac{6}{5}} ]

Therefore, the roots of the equation are:

[ x = \frac{15 \pm \sqrt{249}}{\frac{6}{5}} ] [ x = \frac{25}{3} \pm \frac{5\sqrt{249}}{3} ]

These are the real and imaginary roots of the given quadratic equation.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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