Answer:
[tex]\large\boxed{t=\dfrac{1\pm\sqrt{1-d}}{4}=\dfrac{1}{4}\left(\pm\sqrt{1-d}\right)}[/tex]
Step-by-step explanation:
[tex]\text{The quadratic formula}\\\\ax^2+bx+c=0\\\\\Delta=b^2-4ac\\\\\text{if}\ \Delta>0,\ \text{then}\ x_1=\dfrac{-b-\sqrt\Delta}{2a}\ \text{and}\ x_2=\dfrac{-b+\sqrt\Delta}{2a}\\\\\text{if}\ \Delta=0,\ \text{then}\ x_0=\dfrav{-b}{2a}\\\\\text{if}\ \Delta<0,\ \text{then no real solution}[/tex]
[tex]-16t^2+8t=d\qquad\text{subtract d from both sides}\\\\-16t^2+8t-d=0\qquad\text{change the signs}\\\\16t^2-8t+d=0\\\\a=16,\ b=-8,\ c=d\\\\\Delta=(-8)^2-4(16)(d)=64-64d=64(1-d)\\\\\sqrt\Delta=\sqrt{64(1-d)}=\sqrt{64}\cdot\sqrt{1-d}=8\sqrt{1-d}\\\\t=\dfrac{-(-8)\pm8\sqrt{1-d}}{2(16)}=\dfrac{8\pm8\sqrt{1-d}}{32}=\dfrac{8(1\pm\sqrt{1-d})}{32}\\\\t=\dfrac{1\pm\sqrt{1-d}}{4}[/tex]
