How do you solve #8x-(3x+5)=6x-19 # using the quadratic formula?
It's not a quadratic but a linear equation.
It's a linear equation and quadratic formula cannot be applied.
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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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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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