how do i solve for x in this problem:(x+5)(x-4)=14?

Then x^2 +x -34=0
x=[-1+&-(1+136)^1/2]/2=5.35235 & -6.35235
GOOD LUCK in summer school.

(x + 5)(x - 4) = 14
x(x - 4) + 5(x - 4) 14 = 0
x² - 4x + 5x - 20 = 14
x² + x - 20 = 14
x² + x - 34 = 0

Using the quadratic formula, x = [-b ± √(b² - 4ac)]/(2a), with a = 1, b = 1, c = -34,

x = 5.352 and -6.352 (correct to 3 dp)

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x^2 + x - 20 = 14

x^2 + x - 34 = 0

x = [ - b ± √ ( b ² - 4 a c ) ] / 2 a

x = [ - 1 ± √ ( 1 + 136 ) ] / 2

x = [ - 1 ± √137 ] / 2

(x + 5)(x - 4) = 14
x(x) - x(4) + 5(x) - 5(4) = 14
x^2 - 4x + 5x - 20 - 14 = 0
x^2 + x - 34 = 0
x = [-b ±√(b^2 - 4ac)]/(2a)

a = 1
b = 1
c = -34

x = [-1 ±√(1 + 136)]/2
x = [-1 ±√137]/2

∴ x = [-1 ±√137]/2

Are you sure the equation isn't (x+5)(x-4)= -14? That comes out to a very clean x = 2 or -3.

Expand the left side first:
x^2+x-20=14

Subtract 14 from both sides:
x^2+x-34=0

Use the quadratic equation:

1) x= (-1 + sqrt(1-4(1)(-34)))/(2(1))
x = (-1 + sqrt(137))/2

2) x= (-1 - sqrt(1-4(1)(-34)))/(2(1))
x = (-1- sqrt(137))/2

For the general quadratic equation, for other similar problems, reference:
http://en.wikipedia.org/wiki/Quadratic_equation

This is a quadratic equation.
Multiply the x terms.You will get

x^2+x-20=14. i.e., x^2+x-34=0.
The product value must be -34 and the sum must be +1.

There is no such combination of numbers.The result will be a fractional number.You can find the answer by using the formula given in the following link,

Expand (x+5)(x-4) Use Foil
x^2 - 4x +5x - 20 = 14 Now bring the 14 to the left. It will be subtracted.
x^2 - 4x +5x -20 - 14 =0
x^2 +x - 34 = 0

You have to use the quadratic equation
a = 1
b = 1
c = - 34

x =( -1 +/- sqr(1 - 4(1)(-34)))/2

x = (- 1 +/- sqr(137) )/2 This would likely do as an answer, but you can go further.

x = (- 1 +/- 11.7)/2
x = 5.35 or
x = - 6.35

first subtract 5 from x and 14 then with that result add 4 to both x and the result from before. That will end up with x = something.