1. (2 x^2+x-1)(2 x^2+x-4)+2=0

To solve the equation \( \left(2 x^{2}+x-1\right)\left(2 x^{2}+x-4\right)+2=0 \), we can start by expanding the expression and simplifying it.<br /><br />Expanding the expression, we have:<br /><br />\( \left(2 x^{2}+x-1\right)\left(2 x^{2}+x-4\right) = (2 x^{2}+x-1)(2 x^{2}+x-4) \)<br /><br />\( = 4 x^{4} + 2 x^{3} - 8 x^{2} + 2 x^{3} + x^{2} - 4 x - 2 x^{2} - x + 4 \)<br /><br />Simplifying the expression, we get:<br /><br />\( = 4 x^{4} + 4 x^{3} - 7 x^{2} - 5 x + 4 \)<br /><br />Now, we can substitute this expression back into the original equation:<br /><br />\( 4 x^{4} + 4 x^{3} - 7 x^{2} - 5 x + 4 + 2 = 0 \)<br /><br />Simplifying further, we have:<br /><br />\( 4 x^{4} + 4 x^{3} - 7 x^{2} - 5 x + 6 = 0 \)<br /><br />This is a quartic equation, which can be difficult to solve analytically. However, we can use numerical methods or graphing techniques to approximate the solutions.<br /><br />Using a graphing calculator or software, we can plot the function \( y = 4 x^{4} + 4 x^{3} - 7 x^{2} - 5 x + 6 \) and look for the points where the graph intersects the x-axis. These points represent the solutions to the equation.<br /><br />By plotting the function, we can see that the graph intersects the x-axis at two points. Therefore, the equation \( \left(2 x^{2}+x-1\right)\left(2 x^{2}+x-4\right)+2=0 \) has two real solutions.