How do you solve #8x-(3x+5)=6x-19 # using the quadratic formula?

Answer 1

It's not a quadratic but a linear equation.

It's a linear equation and quadratic formula cannot be applied.

#8x - 3x - 5 = 6x - 19#
#8x - 3x - 6x = 5 - 19#
#x = 14#
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Answer 2

To solve the equation 8x - (3x + 5) = 6x - 19 using the quadratic formula, you first simplify the equation to isolate the variable term. After simplification, you'll have a quadratic equation in the form ax^2 + bx + c = 0, which can then be solved using the quadratic formula:

ax^2 + bx + c = 0

Where a = coefficient of x^2 term, b = coefficient of x term, and c = constant term.

In the given equation: a = 8 - 3 = 5 b = -6 c = 5 - 19 = -14

Then, applying the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

Plug in the values of a, b, and c:

x = (6 ± √((-6)^2 - 4 * 5 * (-14))) / (2 * 5)

Calculate inside the square root first:

b^2 - 4ac = (-6)^2 - 4 * 5 * (-14) = 36 + 280 = 316

Now substitute back into the formula:

x = (6 ± √316) / 10

The roots of the quadratic equation are given by:

x = (6 + √316) / 10 x = (6 - √316) / 10

You can further simplify the roots if required.

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Answer from HIX Tutor

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.

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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