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How do you use the quadratic formula to find both solutions to the quadratic equation #(2y - 3) (y + 1) = 5#?

Answer 1

Savannah Anderson

Multiply out and rearrange into the form #ay^2+by+c = 0# then use the quadratic formula to find:

#y = (1+-sqrt(65))/4#

#5 = (2y-3)(y+1) = 2y^2-y-3#

Subtract #5# from both sides to get:

#2y^2-y-8 = 0#

This is of the form #ay^2+by+c = 0#

with #a=2#, #b=-1# and #c=-8#

This has solutions given by the quadratic formula:

#y = (-b +-sqrt(b^2-4ac))/(2a)#

#=(1+-sqrt((-1)^2-(4xx2xx-8)))/(2*2)#

#=(1+-sqrt(65))/4#

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Answer 2

Jameson Allen

To use the quadratic formula to find both solutions to the quadratic equation (2y - 3)(y + 1) = 5, first expand the left side of the equation and rearrange it into the standard quadratic form ax^2 + bx + c = 0. After that, identify the values of a, b, and c. Then, substitute these values into the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a). Finally, solve for the values of y using the formula.

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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.