Solve the quadratic for x. Algebra 2, Seems pretty simple

Seems simple I just forgot.
(x+2)(x-3 )= 20
I know I can guess and check but I have to show my work

Multiply everything out. Get it all on same side so it is equal to 0. Refactor. Set each factor = 0 or use quadratic formula.

(x + 2)(x - 3) = 20
x^2 -x - 6 = 20
x^2 - x - 26 = 0

This looks hard to do because 6 * 4 =24, not 26...

[-b +/- sqrt(b^2 - 4ac)]/[2a]

[1 +/- sqrt (1 - 4 * 1 * -26)]/[2*1]

0.5 +/- 0.5 * sqrt(105)

First expand (x+2)(x-3) = 20
x² - x -26 = 0 use the quadratic formula {-b ± √ [b² - (4*a*c)]} / 2a

x = [-1 ± √ 105} / 2

(x+2)(x-3) = 20
x^2 - x - 6 = 20
x^2 - x - 26 = 0

-b ± √(b^2 - 4ac) / 2a
-(-1) ± √((-1)^2 - 4(1)(-26)) / 2(1)
1 ± √(1 - (-104)) / 2
1 ± √(1 + 104) / 2
1 ± √105 / 2

1 + √105 / 2 ≈ 1 + 10.247 / 2 ≈ 11.247 / 2 ≈ 5.6235
1 - √105 / 2 ≈ 1 - 10.247 / 2 ≈ / 2 ≈ -9.247 / 2 ≈ -4.6235

(x + 2)(x - 3) = 20

x^2 - x - 6 = 20

x^2 - x = 26

Complete the square"

x^2 - x + 1/4 = 26 + 1/4

(x - 1/2)^2 = 104/4 + 1/4 = 105/4

x - 1/2 = ±√(105/4)

x = 1/2(1 + √105)

x = 1/2(1 - √105)

(x + 2)(x - 3) = 20
=> x² - x - 6 = 20
=> x² - x = 26
=> x² - 2(x)(1/2) + (1/2)² = 26 + (1/2)²
=> (x - 1/2)² = 105/4
=> x - 1/2 = ±√105 / 2
=> x = (1 ± √105)/2