Verified answer

√(x²-3x-10)+√(x²+3x+2)=√(x²+8x+12)

найдем одз

D = 9 + 40 = 49   x12 = (3 +- 7)/2 = 5   -1    (x + 2)(x - 5) >= 0

x ∈ (-∞, -2] U [5, +∞)

D = 9 - 8 = 1   x12 = (-3 +- 1)/2 = -1  -2  (x + 1)(x + 2) >= 0

x ∈ (-∞, -2] U [-1, +∞)

D = 84 - 48 = 16   x12 = (-8 +- 4)/2 = - 6   -2  (x + 2)(x + 6) >= 0

x ∈ (-∞, -6] U [-2, +∞)

общее x∈(-∞,-6] U {-2} U [5,+∞)

√(x²-3x-10)+√(x²+3x+2)=√(x²+8x+12)

√ (x + 2)(x - 5) + √(x + 1)(x + 2) = √(x + 2)(x + 6)

x = -2 корень "убираем" его

√(x - 5) + √(x + 1) = √(x + 6)

x - 5 + 2√(x - 5)√(x + 1) + x + 1 = x + 6

2√ (x - 5)√(x + 1) = 10 - x  (x <= 10 x>=5)

4(x² - 4x - 5) = 100 - 20x + x²

4x² - 16x - 20 = 100 - 20x + x²

3x² + 4x - 120 = 0

D = 16 + 4*3*120 = 1456 = 16*91

x12 = (-4 +- 4√91)/6 = (-2 +- 2√91)/3 ≈ 5.7    -7

x1 =  (-2 - 2√91)/3 <= 5 нет

x2 =   (-2 + 2√91)/3  >=5 и <= 10  корень

ответ х = {  (-2 + 2√91)/3, -2}