Answer:
The discriminant is greater than 0, so there are two real roots.
Explanation:
![\boxed{\begin{minipage}{6.2 cm}\underline{Discriminant}\\\\$b^2-4ac$ \quad when $ax^2+bx+c=0$\\\\If $b^2-4ac > 0 \implies$ two real roots\\If $b^2-4ac=0 \implies$ one real root\\If $b^2-4ac < 0 \implies$ no real roots\\\end{minipage}}](https://img.qammunity.org/2023/formulas/mathematics/college/2o2s5r6vlj5gp1puc4mpid5redpulzdome.png)
Given equation:
![2x^2-9x+2=-1](https://img.qammunity.org/2023/formulas/mathematics/college/jgr54m95mcsngerchs7lia6x05co7w4dtn.png)
Add 1 to both sides of the equation so that the equation equals zero:
![\implies 2x^2-9x+2+1=-1+1](https://img.qammunity.org/2023/formulas/mathematics/college/bwsq6bujazya1pnpk7hoxkftfdj5kamu23.png)
![\implies 2x^2-9x+3=0](https://img.qammunity.org/2023/formulas/mathematics/college/jt64xom91a8pi36tjf1zgjd99r2knxpsx1.png)
Compare the equation with ax²+bx+c=0:
Substitute the values of a, b and c into the discriminant formula and solve:
![\begin{aligned}\implies b^2-4ac&=(-9)^2-4(2)(3)\\&=81-4(2)(3)\\&=81-8(3)\\&=81-24\\&=57\end{aligned}](https://img.qammunity.org/2023/formulas/mathematics/college/xsh6eae6dsh482kdo5be6m7jmg0inho109.png)
As 57 > 0, the discriminant is greater than zero, so there are two real roots.