Show algebraically that for all real x, 5+10x−3x^2≤40/3?

5 + 10x - 3x²
= 5 - (3x² - 10x)
= 5 - 3[x² - (10/3)x]
= 5 - 3[x² - 2(5/3)x + (5/3)²] + 3*(5/3)²
= 5 - 3[x - (5/3)]² + 25/3
= (40/3) - 3[x - (5/3)]²

For all real x, -3[x - (5/3)]² ≤ 0
Then, 5 + 10x - 3x² = (40/3) - 3[x - (5/3)]² ≤ 40/3

5+10x−3x^2≤40/3 ------- (1)

First, solve 5+10x−3x^2 = 40/3
multiply both sides by 3
15+30x-9x^2 = 40
-9x^2+30x+15-40 = 0
9x^2-30x-15+40 = 0
9x^2-30x+25 = 0

This equation is of form ax^2+bx+c=0
a = 9 b = -30 c = 25
x=[-b+/-sqrt(b^2-4ac)]/2a]
x=[30 +/-sqrt(-30^2-4(9)(25)]/(2)(9)
discriminant is b^2-4ac =0
x=[30 +√(0)] / (2)(9)
x=[30 -√(0)] / (2)(9)
x=[30+0] / 18
x=[30-0] / 18

x=15/9

------- ------ ------------
-∞.........15/9......∞

Consider the intervals (-∞,15/9) and (15/9,∞)

Choose any one point from each interval and test (1)
(-∞,15/9) : choose x=1
5+10x-3x^2 ≤ 40/3
12 ≤ 40/3 (true)

(15/9,∞) : choose x=2
5+10x-3x^2 ≤ 40/3
5+20-12 = 5 < = 40/3 (true)

5+10x−3x^2≤40/3 is true for all x
-∞ < x < ∞

5 + 10x − 3x^2 ≤ 40/3

15 + 30x - 9x^2 ≤ 40

- 25 + 30x - 9x^2 ≤ 0

- [ 9x^2 + 30x + 25] ≤ 0

- ( 3x - 5)^2 ≤ 0

For all values of x, ( 3x - 5)^2 > 0

Hence

- ( 3x - 5)^2 ≤ 0 is valid for all values of x