Решите третье умоляю

$x^{2} = -\dfrac{x}{4} + 0,75$

$x^{2} + \dfrac{x}{4} - 0,75 = 0 | \cdotp 4$

$4x^{2} + x - 3 = 0$

$a = 4 ; b = 1; c = -3$

$D = b^{2} - 4ac = 1^{2} - 4 \cdotp 4 \cdotp (-3) = 1 + 48 = 49$

$x_{1} = \dfrac{-b + \sqrt{D}}{2a} = \dfrac{-1 + \sqrt{49}}{2 \cdotp 4} = \dfrac{-1 + 7}{8} = \dfrac{6}{8} = \dfrac{3}{4}$

$x_{2} = \dfrac{-b - \sqrt{D}}{2a} = \dfrac{-1 - \sqrt{49}}{2 \cdotp 4} = \dfrac{-1 - 7}{8} = -1$

Ответ: Г