Найдите значение p, если корни уравнения 6х^2+3x - p = 0; если корни уравнения удовлетворяют условию x1 * x2^4 +x2 *x1^4=63/8 помогите плиз или хотя бы скажите как решать....

 6х^2+3x - p = 0 D = 9 +4*6*p = 9 + 24p x = [-3 +- (9 + 24p)^0.5]/12   x1 * x2^4 +x2 *x1^4=63/8 [latex]\frac{-3+\sqrt{9+24p}}{12}\\ \frac{-3+\sqrt{9+24p}}{12}*(\frac{-3-\sqrt{9+24p}}{12})^{4} + \frac{-3-\sqrt{9+24p}}{12}*(\frac{-3+\sqrt{9+24p}}{12})^{4}=\\ \frac{-3+\sqrt{9+24p}}{12}*\frac{-3-\sqrt{9+24p}}{12}*[(\frac{-3-\sqrt{9+24p}}{12})^{3} + (\frac{-3+\sqrt{9+24p})^{}}{12})^{3}]=\\ \frac{(-3)^{2}-(\sqrt{9+24p})^{2}}{144}*[(\frac{-3-\sqrt{9+24p}}{12} + \frac{-3+\sqrt{9+24p}}{12})*\\ *((\frac{-3-\sqrt{9+24p}}{12})^{2} + (\frac{-3+\sqrt{9+24p})}{12})^{2} - \frac{-3-\sqrt{9+24p}}{12}*\frac{-3+\sqrt{}}{12}] [/latex]   [latex]\frac{9-9-24p}{144}*[(\frac{-6}{12})*(\frac{9+9+24p+9+9+24p}{144} - \frac{9-9-24p}{144})]=\\ =-\frac{p}{6}*[(-\frac{1}{2})*(\frac{36+48p}{144} + \frac{24p}{144})]=\\ =\frac{p}{12}*\frac{36p+72p}{144} = \frac{63}{8}\\ 36p+72p^{2} = 12*63*18\\ 36p+72p^{2} = 13608\\ p+2p^{2} = 378\\ 2p^{2} +p - 378 = 0\\ D = 1 + 4*2*378 = 3025 = 55^{2}\\ p = \frac{-1 ^{+}_{-} 55}{4}\\ p_{1} = 13,5\\ p_{2} = -14[/latex]