How do you use the quadratic formula to find both solutions to the quadratic equation #2x^2+3x-5=0#?
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In this case (a + b + c = 0), you don't need using the formula. Use the shortcut. One real root is (1) and the other is (c/a = -5/2)
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To use the quadratic formula to find both solutions to the quadratic equation 2x^2 + 3x - 5 = 0, you first identify the coefficients a, b, and c in the general form of the quadratic equation (ax^2 + bx + c = 0). Then, substitute these values into the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For the equation 2x^2 + 3x - 5 = 0:
a = 2 b = 3 c = -5
Substitute these values into the quadratic formula:
x = (-(3) ± √((3)^2 - 4(2)(-5))) / (2(2))
Simplify under the square root:
= (-(3) ± √(9 + 40)) / (4)
= (-(3) ± √49) / (4)
= (-(3) ± 7) / (4)
Now, you have two possible solutions:
- x = (-3 + 7) / 4 = 4 / 4 = 1
- x = (-3 - 7) / 4 = -10 / 4 = -5/2
So, the solutions to the quadratic equation 2x^2 + 3x - 5 = 0 are x = 1 and x = -5/2.
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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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