Best Answer

It is a quadratic equation in the form of y2-4y-5 = 0 and will have two solutions:

When factorised:

(y-5)(y+1) = 0

Therefore: y = 5 or y = -1

Q: What is y squared minus 4y minus 5?

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it depends,if y = 0 the answer is -12 if y = 5 then the answer is 368

y2-4y-32 = (y+4)(y-8) when factored

2

(2y+1)(y−5)

Perhaps, using the Quadratic Equation...y=x2+5x-6Solve for x:x2+5x+(-6-y)=0Using the Quadratic Equation:x = .5 *(-5 (+-) (25 + (24+4y)).5) = -2.5 (+-) .5*(49+4y).5Substitute y for x, x for y:y = -2.5 (+-) .5*(49+4x).5

Related questions

it depends,if y = 0 the answer is -12 if y = 5 then the answer is 368

4(y - 3)(y + 3)

y2-4y-32 = (y+4)(y-8) when factored

-6 + 3x - y - 4x + 4y + 5 (combine like terms) = (-6 + 5) + (3x - 4x) + (-y + 4y) = -1 - x + 3yHope this helps :)

2

First, solve for zero y2 - 4y > 5 y2 - 4y - 5 > 0 (y - 5)(y + 1) > 0 So where y = 5 and y = -1, the inequality solves for zero. We know that because the term with the highest power of y has a positive coefficient, that this is a parabola which has a limit in the negative, and is infinite in the positive. That means y is less than negative one or greater than 5. {x | x < -1 OR x > 5, x ∈ ℜ}

7(3y-2)-(-4y+9)

Perhaps, using the Quadratic Equation...y=x2+5x-6Solve for x:x2+5x+(-6-y)=0Using the Quadratic Equation:x = .5 *(-5 (+-) (25 + (24+4y)).5) = -2.5 (+-) .5*(49+4y).5Substitute y for x, x for y:y = -2.5 (+-) .5*(49+4x).5

(2y+1)(y−5)

If: x -y = 2 then x^2 = y^2 +4y +4 If: x^2 -4y^2 = 5 then x^2 = 4y^2 +5 So: 4y^2 +5 = y^2 +4y +4 Transposing terms: 3y^2 -4y +1 = 0 Factorizing the above: (3y -1)(y -1) = 0 meaning y = 1/3 or y = 1 Substitution into the original linear equation intersections are at: (7/3, 1/3) and (3, 1)

If: x -y = 2 then x^2 = (2+y)^2 => 4+4y+y^2 If: x^2 -4y^2 = 5 then x^2 = 5+4y^2 So: 5+4y^2 = 4+4y+y^2 Transposing terms: 3y^2 -4y +1 = 0 Factorizing the above: (3y-1)(y-1) = 0 meaning y = 1/3 or y = 1 By substitution contacts are made at: (7/3, 1/3) and (3, 1)

If: x -y = 2 then x = 2 +y If: x^2 -4y^2 = 5 then (2 +y)^2 -4y^2 = 5 So: 4+4y +y^2 -4y^2 = 5 Collecting like terms: 4y -3y^2 -1 = 0 Solving the above quadratic equation: y = 1 or y = 1/3 Therefore by substitution the points of contact are at: (3, 1) and (7/3, 1/3)