Question

Plz factorise this
10t^2-10t+2.5

Answer 1

$10 {t}^{2} - 10t + 2.5 = 0 \\ 20 {t}^{2} - 20t + 5 = 0 \\ 4 {t}^{2} - 4t + 1 = 0 \\ {(2t)}^{2} - 2 \times 2t \times 1 + {1}^{2} = 0 \\ {(t - 2)}^{2} = 0$
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Answer 2

Given Equation is 10t^2 - 10t + 2.5

= > 10t^2 - 2 * t * 0.5 * 10 + 10 * 0.5 * 0.5

= > 10(t^2 - 2 * t * 0.5 + 0.5 * 0.5)

We know that (a - b)^2 = a^2 + b^2 - 2ab.

= > 10(t - 0.5)^2.

Hope this helps!