Question

Solve for t d=−16t^2+8t

t=8±1−d−−−−√ t=14±1−d√4 t=12±81−d−−−−√ t=14±81−d−−−−√

Answer 1

`\large\boxed{t=(1\pm√(1-d))/(4)=(1)/(4)\left(\pm√(1-d)\right)}`

Explanation:

`\text{The quadratic formula}\\\\ax^2+bx+c=0\\\\\Delta=b^2-4ac\\\\\text{if}\ \Delta>0,\ \text{then}\ x_1=(-b-\sqrt\Delta)/(2a)\ \text{and}\ x_2=(-b+\sqrt\Delta)/(2a)\\\\\text{if}\ \Delta=0,\ \text{then}\ x_0=\dfrav{-b}{2a}\\\\\text{if}\ \Delta<0,\ \text{then no real solution}`

`-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=√(64(1-d))=√(64)\cdot√(1-d)=8√(1-d)\\\\t=(-(-8)\pm8√(1-d))/(2(16))=(8\pm8√(1-d))/(32)=(8(1\pm√(1-d)))/(32)\\\\t=(1\pm√(1-d))/(4)`