Write a quadratic equation in standard form that has two solutions, 6, and -10.?

Question

The leading coefecient must be one

Answer 1

x^2+4x-60

Answer 2

The solutions are 6 and -10.
So it is
(x - 6)(x + 10) = 0
x ^ 2 + 4x - 60

Answer 3

solutions are x= 6..means x-6=0
x=-10 means x+10=0
so the eqation must have (x-6) and (x+10)
so the qudratic eqation has to be the product of (x-6) and (x+10)
expansion of which gives
x^2-6x+10x-60
i.e x^2-4x-60

Answer 4

It is: x^2 + 4x - 60 = 0.

Answer 5

permit the no of mugs offered be n cost/mug=40 8/n no of mugs offered=n-2 advertising cost= no of mugs offered(n-2) prolonged by way of the S.P. of one mug{(40 8/n)+3} yet that's additionally equivalent to the S.P. that's comparable to cost +earnings (40 8+22) and so equating (n-2)(40 8/n +3)=40 8+22=70 (n-2)(40 8+3n)=70n 48n-6n-ninety six+3n^2=70n rewriting interior the std Q.E. variety 3n^2-28n-ninety six=0 3n^2-36n+8n-ninety six=0 3n(n-12)+8(n-12)=0 if the product is 0 the two component is 0 so n=12 or-8/3 which isn't suited cost cost of the mug=40 8/12=$4 advertising cost=70/10=$7 earnings /mug=$3 no of mugs offered=n and a pair of have been given broken so saleable mugs is two decrease than the no offered=n-2 advertising cost according to mug=gross sales/no of mugs offered=70/(n-2) gross sales=C.P.+earnings=40 8+22=$70

Answer 6

it is (x-6)(x+10)
=x^2+4x-60

Answer 7

y=(x-6)(x+10)
y=x^2-6x+10x-60
y=x^2+4x-60

Answer 8

x^+4x-60