How do you solve the equation x^{2}+7x=18?

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How do you solve the equation x^{2}+7x=18?

Thank you!

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Subtract 18 from both sides to get the equation set to 0:

x^{2} + 7x - 18 = 0

Now factor. Given the negative sign in front of the 18, you will have (x + ) (x - ) as your factors, with the larger one being positive. Now, which factors of 18 can be combined to get a result of 7? I'll leave that for you.

x^{2}+7x-18=0 is an eqaution of a quadratic form ax^{2}+bx +c = 0

For the above equation a=1, b= 7, and c = -18

to test how many roots are for this equation compute b^{2 }- 4 ac, (7)^{2}-4(1)(-18) = 121

Since b2- 4ac > 0 or positive the equation has two roots, please note that if b2-4ac < 0 or negative, the equation has no roots, and if b2-4ac = 0, the equation has one root and therefore one solution.

The root are found by x = (- b ± √b^{2}-4ac ) / 2a , and x = (-7 ± √121 ) / 2 (1) = ( -7 ± 11) / 2

The two roots are x1 = ( -7 - 11) /2 = -18 /2 = - 9

and x2 = ( -7 + 11) / 2 = 4 /2 = 2

To check you answer substitute each x value in the equation to get zero

for x1 = -9 , then (-9)^{2}+7(-9)-18 = 81 -63 -18 = 0

for x2 = 2, then (2)^{2}+ 7 (2) -18 = 4 + 14 - 18 = 0

Threfore both values of x are valid for the equations, and the equation has two roots x = -9 , and x = 2