leading to 2x2+x−15=0 with a valid attempt to solve to obtain 2 values for x M1 x=25,−3 A1 For both Mark final answer for this equation, A0 if - 3 is rejected x2+x−10=−5
leading to 2x2+x−5=0 M1 Must be a correct 3-term quadratic equated to zero x=4−1±41 oe 2 Dep M1 for a valid method of solution to obtain 2 values for x. Mark final answer for this equation, A0 if negative root is rejected Denominator of final answer must be positive