Given the following quadratic equation:
![\text{ -2y}^2\text{ + y + 7 = 0}](https://img.qammunity.org/2023/formulas/mathematics/college/o75tbaf6q067uvbtoy48dq058hpe2cw6wj.png)
The given equation is already in standard form a^2y + by + c = 0. Thus, we can say that a, b and c values are the constants of each term.
Therefore,
a = -2
b = 1
c = 0
To be able to determine the solution, we will be using the following formula:
![\text{ y = }\frac{-b\text{ }\pm\text{ }\sqrt[]{b^2-4ac}}{2a}](https://img.qammunity.org/2023/formulas/mathematics/college/gtesnq88vuusz9hz3zoe7hzd97qwid6ida.png)
Plugging in the values of a, b and c, we get:
![\text{ y = }\frac{-1\text{ }\pm\text{ }\sqrt[]{(1)^2-4(-2)(0)}}{2(-2)}](https://img.qammunity.org/2023/formulas/mathematics/college/osd0reszhhgc97zr3y2kwwbrv1cnji13li.png)
![\text{ y = }\frac{-1\text{ }\pm\text{ }\sqrt[]{1-0)}}{-4}](https://img.qammunity.org/2023/formulas/mathematics/college/spo0vbvboqa9in59eyaf6teddz6yzyewar.png)
![\text{ y = }\frac{-1\text{ }\pm\text{ }\sqrt[]{1}}{-4}](https://img.qammunity.org/2023/formulas/mathematics/college/qm13cx882uol7vyxxwtkj40j6c9m2tdtbz.png)
![\text{ y = }\frac{-1\text{ }\pm\text{1}}{-4}](https://img.qammunity.org/2023/formulas/mathematics/college/xvj0bsvuwkddih6t8msjblob5sbvox8dbu.png)
![y_1\text{ = }\frac{-1\text{ + 1}}{-4}\text{ = }(0)/(-4)\text{ = 0}](https://img.qammunity.org/2023/formulas/mathematics/college/cmtn64xhm22ata82xegsvie2qlzyrgnivo.png)
![y_2\text{ = }\frac{-1\text{ - 1}}{-4}\text{ = }(-2)/(-4)\text{ = }(2)/(4)\text{ = }(1)/(2)\text{ or 0.5}](https://img.qammunity.org/2023/formulas/mathematics/college/99bk26k2ns0yjd1r0txo91k09cvgewlbs5.png)
Therefore, the solutions of the given quadratic equation are y = 0 and y = 1/2 or 0.5. It has two different solutions.