Решите уравнение пж (2х²+3х-1)²-10х²-15+9=0

(2х² + 3х - 1)² - 10х² - 15х + 9 = 0
(2х² + 3х - 1)² - 5(2х² + 3х) + 9 = 0

(2х² + 3х) = y

(y - 1)² - 5y + 9 = 0
y² - 2y + 1 - 5y + 9 = 0
y² - 7y + 10 = 0
D = b² - 4ac = 49 - 4 * 1 * 10 = 9
$y_{1} = \frac{-b+ \sqrt{D} }{2} = \frac{7+3}{2} = 5 y_{2} = \frac{-b- \sqrt{D} }{2} = \frac{7 -3 }{2} = 2$

(2х² + 3х) = 5
2х² + 3х - 5 = 0
D = b² - 4ac = 9 - 4 * 2 * (-5) = 49
$x_{1} = \frac{-b+ \sqrt{D} }{2} = \frac{-3+7}{2} = 2 x_{2} = \frac{-b- \sqrt{D} }{2} = \frac{-3 -7}{2} = -5$

(2х² + 3х) = 2
2х² + 3х - 2 = 0
D = b² - 4ac = 9 - 4 * 2 * (-2) = 25
$x_{1} = \frac{-b+ \sqrt{D} }{2} = \frac{-3+5}{2} = 1 x_{2} = \frac{-b- \sqrt{D} }{2} = \frac{-3 -5}{2} = -4$

Ответ: 2; -5; 1; -4.