[tex]\displaystyle\bf\\\left \{ {{y = 2 {x}^{2} + 3x - 5 } \atop {y = 2x + 1 }} \right. \\ \\ 2 {x}^{2} + 3x - 5 = 2x + 1 \\ 2 {x}^{2} + 3x - 2x - 5 - 1 = 0 \\ 2 {x}^{2} + x - 6 = 0 \\ a =2 \\ b = 1 \\ c = - 6 \\ D = {b}^{2} - 4ac = 1 {}^{2} - 4 \times 2 \times ( - 6) = \\ = 1 + 48 = 49 \\ x_{1} = \frac{ - 1 - 7}{2 \times 2} = - \frac{8}{4} = - 2\\ x_{2} = \frac{ - 1 + 7}{2 \times 2} = \frac{6}{4} = 1.5 \\ \\ y_{1} = 2 \times ( - 2) + 1 = - 4 + 1 = - 3 \\ y_{2} = 2 \times 1.5 + 1 = 3 + 1 = 4[/tex]

Ответ: ( - 2 ; - 3 ) и ( 1,5 ; 4 )