# Rearranging algebraic formula when subject appears twice and is squared

I am unsure how to tackle making $$t$$ the subject of this formula

$$D = ut + kt^2$$

Because it appears twice and is squared, does this mean I have to factorise quadratically? I have tried but how is this possible when you only have algebra coefficients? Also surely if it was factorised into 2 brackets then you still wouldn't be able to single out the subject onto one side, so I guess you would have to factorise it into one bracket?

Please can someone make sense of this for me, explaining simply? Thankyou

• Sorry; what is it you're trying to do? Jun 3 '19 at 19:50
• Complete the squares. $kt^2+ut=k(t^2+\frac ukt)=k\left(t^2+\frac uk t+\frac {u^2}{4k^2}-\frac {u^2}{4k^2}\right)=k(t+\frac {u}{2k})^2-k\times \frac {u^2}{4k^2}$
– lulu
Jun 3 '19 at 19:52
• You could also use the quadratic formula after subtracting D on both sides. Jun 3 '19 at 19:53
• Yes. As has been remarked, this is just the same as using the quadratic formula. In fact, this is how the quadratic formula is demonstrated.
– lulu
Jun 3 '19 at 20:02
• Well, at a minimum your expression should allow both signs for the square root. But let's check your expression using the quadratic formula. Rearrange the original to get $kt^2+ut-D=0$. The quadratic formula then tells us that $t=\frac {-u\pm \sqrt {u^2+4kD}}{2k}$. Can you check that this is the same as your expression (other than the signs of the square root)?
– lulu
Jun 3 '19 at 20:16