Відповідь:

[tex]x = 0\\x = \frac{1}{2}\\ x = \frac{ 1 + \sqrt{57} }{4} \\x = \frac{1 - \sqrt{57}}{4}[/tex]

Покрокове пояснення:

[tex]2x^2 -x -5 )^2 +3(2x^2-x-5)-10 = 0\\2x^2-x-5 = t\\\\t^2 + 3t - 10 = 0\\\\t^2+5t-2t- 10 = 0\\t*(t+5)-2(t+5)=0\\(t+5)*(t-2)=0 = > \left \{ {{t + 5 = 0} \atop {t - 2 =0}} \right = > \left \{ {{t = -5} \atop {t = 2}} \right. \\\\2x^2 - x - 5 = -5\\2x^2-x-5 = 2\\2x^2 -x -5= -5\\2x^2 -x = 0\\x*(2x-1) = 0 = > \left \{ {{x=0} \atop {2x-1 = 0}} \right. = > \left \{ {{x = 0} \atop {x = \frac{1}{2} }} \right. \\\\[/tex]

[tex]2x^2 - x -5 -2 = 0\\2x^2-x-7 = 0\\a = 2, b = -1, c = -7\\x = \frac{-b += \sqrt[2]{b^2-4ac} }{2a} = > x = \frac{-(-1) +- \sqrt{(-1)^2 - 4 * 2 * (-7)} }{2*2\\}\\ x = \frac{1 +- \sqrt{1 - 4 * 2 * (-7)} }{4} = > x = \frac{1 +- \sqrt{57} }{4} \\\\[/tex]

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