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)  $x = 4\pm2\sqrt{10}$

This is the same as saying $x = 4+2\sqrt{10} \ \ \text{ or } \ \ x = 4-2\sqrt{10}$

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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

$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x = \frac{-(-8)\pm\sqrt{(-8)^2-4(1)(-24)}}{2(1)}\\\\x = \frac{8\pm\sqrt{160}}{2}\\\\x = \frac{8\pm\sqrt{16*10}}{2}\\\\x = \frac{8\pm\sqrt{16}*\sqrt{10}}{2}\\\\x = \frac{8\pm4*\sqrt{10}}{2}\\\\x = \frac{2(4\pm2\sqrt{10})}{2}\\\\x = 4\pm2\sqrt{10}$

This shows the answer is choice C.

The plus minus breaks down that equation these two solutions

$x = 4+2\sqrt{10} \ \ \text{ or } \ \ x = 4-2\sqrt{10}$