Question

PLEASE PLEASE HELP ME OUT ITS URGENT

What are the solutions to the quadratic equation?

Answer 1

Answer:

x^2+13=8x+37

x^2+13-8x-37=0

x^2-24-8x=0

x^2-8x-24=0

x= -(-8)plus or minus [the square root of (-8)^2-4*1(-24)]/2*1

x= 8 plus or minus [the square root of 64+96]/2

x= 8 plus or minus [the square root of 160]/2

x= 8 plus or minus 4 [the square root of 10]/2

x=8+4 [the square root of 10]/2
x=8-4 [the square root of 10]/2

or simplified,

x=4+2 [the square root of 10]
x=4-2 [the square root of 10]

So your answer is C. x=4 plus or minus 2 [the square root of 10]

Answer 2

Answer: Choice C)
`x = 4\pm2√(10)`

This is the same as saying
`x = 4+2√(10) \ \ \text{ or } \ \ x = 4-2√(10)`

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Step-by-step explanation:

First get everything to one side

x^2 + 13 = 8x + 37

x^2 + 13 - 8x - 37 = 0

x^2 - 8x - 24 = 0

Here we have a quadratic in the form ax^2+bx+c = 0

Note how a = 1, b = -8, c = -24

Those values are plugged into the quadratic formula below

`x = (-b\pm√(b^2-4ac))/(2a)\\\\x = (-(-8)\pm√((-8)^2-4(1)(-24)))/(2(1))\\\\x = (8\pm√(160))/(2)\\\\x = (8\pm√(16*10))/(2)\\\\x = (8\pm√(16)*√(10))/(2)\\\\x = (8\pm4*√(10))/(2)\\\\x = (2(4\pm2√(10)))/(2)\\\\x = 4\pm2√(10)`

This shows the answer is choice C.

The plus minus breaks down that equation these two solutions

`x = 4+2√(10) \ \ \text{ or } \ \ x = 4-2√(10)`