Question

Q.16 please With steps

Answer 1

The equations

$2x^2+9x-5=0$

and

$2\left(t-\dfrac12\right)^2+9\left(t-\dfrac12\right)-5=0$

are really the same, we're just setting $x=t-\dfrac12$, or $t=x+\dfrac12$.

$2x^2+9x-5=(2x-1)(x+5)=0\implies x=\dfrac12,x=-5$

So we get

$t=\dfrac12+\dfrac12=1,t=-5+\dfrac12=-\dfrac92$