(x+4)(x+5)=20

To solve the equation (x+4)(x+5)=20, we will expand the left side of the equation and then solve for x.

Expanding the left side:

(x+4)(x+5) = x * x + x * 5 + 4 * x + 4 * 5 = x^2 + 5x + 4x + 20 = x^2 + 9x + 20

Now we have the equation x^2 + 9x + 20 = 20

Subtracting 20 from both sides:

x^2 + 9x + 20 - 20 = 20 - 20 x^2 + 9x = 0

Now we have the quadratic equation x^2 + 9x = 0. We can solve this by factoring.

Factoring out x:

x(x + 9) = 0

Setting each factor equal to zero:

x = 0 or x + 9 = 0

If x = 0, then the equation (x+4)(x+5) = 20 is not satisfied, because (0+4)(0+5) = 4 * 5 = 20, which is not equal to 20.

If x + 9 = 0, then x = -9. We can check if this value of x satisfies the equation:

(-9+4)(-9+5) = -5 * -4 = 20

Therefore, the solution to the equation (x+4)(x+5)=20 is x = -9.