To determine if a given value is a solution to the quadratic equation −2x^2 − 3x + 2 = 0, we substitute it in for x and evaluate the equation.
For x = −34:
−2(−34)^2 − 3(−34) + 2 = −2(1156) + 102 + 2 = −2312 + 102 + 2 = −2208 + 2 = −2206
For x = −12:
−2(−12)^2 − 3(−12) + 2 = −2(144) + 36 + 2 = −288 + 36 + 2 = −250
For x = 12:
−2(12)^2 − 3(12) + 2 = −2(144) + 36 + 2 = −288 + 36 + 2 = −250
None of the given values result in the equation being true. Therefore, none of the values are in the solution set. The statement "Only x= is in the solution set" is incorrect.