(x-5)(x+19)=585
(x-5)(x+19) = x^2 + 14x - 95
second degree equation (y = a*x^2 + b*x +c), must use Delta = b^2 - 4*a*c to solve it
*** Solution ***
Delta = (14)^2 - 4*(1)*(-95) = 596 > 0
two solutions:
x1 = (-b-sqrt(Delta))/(2*a)
x2 = (-b+sqrt(Delta))/(2*a)
Numeric Application :
x1 = -19
x2 = +5
Joe ignored the 585 on the Right Side
(x-5)(x+19)=585
x^2 + 14x - 95 = 585
x^2 + 14x + .... = 680
x^2 + 14x + 49 = 680 + 49 using completing the square because of the nice numbers
(x+7)^2 = 729
x+7 = ±27
x = 20 or x = -34
which could have been found also by factoring
(x-20)(x+34) = 0
etc.
true i am sorry for that !
thanks Reiny for correcting :)
To solve this equation, we can use the method of factoring. The first step is to simplify the equation by applying the distributive property.
(x - 5)(x + 19) = 585
Now, let's expand the left side of the equation by multiplying the terms:
x(x) + x(19) - 5(x) - 5(19) = 585
x^2 + 19x - 5x - 95 = 585
Combining like terms, we get:
x^2 + 14x - 95 = 585
Now, let's bring all the terms to one side of the equation by subtracting 585 from both sides:
x^2 + 14x - 95 - 585 = 0
x^2 + 14x - 680 = 0
We now have a quadratic equation in the form of ax^2 + bx + c = 0 where a = 1, b = 14, and c = -680. To solve this quadratic equation, we can use factoring, completing the square, or the quadratic formula. In this case, since factoring is possible, let's try factoring:
The product of the constant term (-680) and the coefficient of x (14) is -9520. We need to find two numbers that multiply to -9520 and add up to 14.
After some trial and error, we can find that the numbers are 40 and -238. Adding them gives us 40 + (-238) = -198, and multiplying them gives us 40 * (-238) = -9520.
Now, let's rewrite the equation factoring the quadratic:
(x + 40)(x - 238) = 0
Now we set each factor equal to zero and solve for x:
x + 40 = 0 --> x = -40
x - 238 = 0 --> x = 238
So the solutions to the equation (x - 5)(x + 19) = 585 are x = -40 and x = 238.