Если x1 и x2-корни уравнения 2x^2+3x-4=0, то значение выражения x1^4+x2^4 равно ?

По теореме Виета
$2x^2+3x-4=0\\ x_{1}+x_{2}=-\frac{3}{2}\\ x_{1}x_{2}=-2\\\\ (x_{1}+x_{2})^4=x_{1}^4+4x_{1}x_{2}^3+6x_{1}^2x_{2}^2+4x_{1}^3x_{2}+x_{2}^4=\frac{81}{16}\\ 4x_{1}x_{2}(x_{2}^2+x_{1}^2)+6x_{1}^2x_{2}^2=\\ x_{1}^2+x_{2}^2=(x_{1}+x_{2})^2-2x_{1}x_{2}=\frac{9}{4}+2*2=\frac{25}{4}\\ 4x_{1}x_{2}(x_{2}^2+x_{1}^2)+6x_{1}^2x_{2}^2=4*-2(\frac{25}{4}) +6*4 = -50+24 = -26\\ \\ x_{1}^4+x_{2}^4=\frac{81}{16}+26=\frac{497}{16}$