Question

Factorise 8n2 -4n-180​

Answer 1

Answer:

82−4−180

8n^{2}-4n-1808n2−4n−180

4(22−−45)

4(2n^{2}-n-45)4(2n2−n−45)

2

Use the sum-product pattern

4(22−−45)

4(2n^{2}{\color{#c92786}{-n}}-45)4(2n2−n−45)

4(22+9−10−45)

4(2n^{2}+{\color{#c92786}{9n}}{\color{#c92786}{-10n}}-45)4(2n2+9n−10n−45)

3

Common factor from the two pairs

4(22+9−10−45)

4(2n^{2}+9n-10n-45)4(2n2+9n−10n−45)

4((2+9)−5(2+9))

4(n(2n+9)-5(2n+9))4(n(2n+9)−5(2n+9))

4

Rewrite in factored form

4((2+9)−5(2+9))

4(n(2n+9)-5(2n+9))4(n(2n+9)−5(2n+9))

4(−5)(2+9)

4(n-5)(2n+9)4(n−5)(2n+9)

Solution

4(−5)(2+9)

Answer 2

Answer:

8n²-4n-180

comparing it with ax²+bx+c=0,we get a=8,b=-4,c=-180

D=b²-4ac

D=(-4)²-4(8)(-180)

D=16-32-180

D=196

by quadratic formula,

x=-b+√D÷2a = -(-4)+√196÷2(8) = 4+14÷16 = 18÷16 =9/8

x=-b-√D÷2a = -(-4)-√196÷2(8) = 4-14÷16 = -10÷16 = -5/8

Step-by-step explanation:

thnks

please mark me as brainlist