Someone please helppp

k = 5/6

Step-by-step explanation:

First, we can make this have the form of a quadratic function, or ax²+bx+c=0. To do this, we can first subtract 10x from both sides to get

(2k+1)x²-8x=-6

Next, we can add 6 to both sides, resulting in

(2k+1)x²-8x+6 = 0

For a quadratic function of form ax²+bx+c=0, we can see that a=2k+1, b=-8, and c=6. We can then apply the quadratic equation, or

x= (-b ± √(b²-4ac))/(2a) to get our roots to be

x= (8 ± √(64-4(6)(2k+1)))/(2*(2k+1))

=  (8 ± √(64-(48k+24)))/(4k+2)

=  (8 ± √(40-48k))/(4k+2)

For the roots to be equal, we must have the two roots equal to each other. We can write this as

(8 + √(40-48k))/(4k+2) = (8 - √(40-48k))/(4k+2)

multiply both sides by (4k+2) to remove the denominator

8+√(40-48k) = 8 - √(40-48k)

subtract 8 from both sides to isolate the square roots

√(40-48k) = - √(40-48k)

The only number that is equal to its negative self (and is real) is 0. Therefore, √(40-48k) = - √(40-48k) = 0, so we have

√(40-48k) = 0

square both sides to remove the square root

40-48k = 0

add 48k to both sides to isolate the k and its coefficient

40 = 48k

divide both sides by 48 to isolate k

k = 40/48 = 5/6

okay wait before i answer, is this for your points on a graph?

damm bro the struggle

step-by-step explanation: