Best solutions to - "How do you solve 10t2−29t=−10?"

High School
Algebra
Algebra I
Quadratic function and equation

Zugdichte2r

Answered question

2022-01-21

How do you solve

10 t^{2} - 29 t = - 10

Answer & Explanation

fionaluvsyou0x

Beginner2022-01-22Added 11 answers

Step 1
$10 t^{2} - 29 t + 10 = 0$
you can solve it by 2 methods, the quadratic formula or splitting the middle term. I am doing it by the latter one.
the product of roots is 100 and the sum is -29
the suitable pair will be -25 and -4
therefore,
$10 t^{2} - 25 t - 4 t + 10 = 0$
$5 t (2 t - 5) - 2 (2 t - 5) = 0$
$(5 t - 2) (2 t - 5) =)$
hich implies that
$t = \frac{2}{5}$ or $\frac{5}{2}$

Darian Sosa

Beginner2022-01-23Added 12 answers

Step 1
Here,
$10 t^{2} - 29 t = - 10$
$10 t^{2} - 29 t + 10 = 0$
$10 t^{2} - 25 t - 4 t + 10 = 0$
$5 t (2 t - 5) - 2 (2 t - 5) = 0$
$(2 t - 5) (5 t - 2) = 0$
$2 t - 5 = 0$ or $5 t - 2 = 0$
$2 t = 5$ or $5 t = 2$
$t = \frac{5}{2}$ or $t = \frac{2}{5}$
or
$10 t^{2} - 29 t + 10 = 0$
Comparing with $a t^{2} + b t + c = 0$
$a = 10, b = - 29$ and $c = 10$
So,
$t = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

$∴ t = \frac{29 \pm \sqrt{29^{2} - 4 (10) (10)}}{2 \times 10}$
$∴ t = \frac{29 \pm \sqrt{841 - 400}}{20}$
$∴ t = \frac{29 \pm 21}{20}$
$∴ t = \frac{29 + 21}{20}$ or $\frac{29 - 21}{20}$
$∴ t = \frac{50}{20}$ or $t = \frac{8}{20}$
$∴ t = \frac{5}{2}$ or $t = \frac{2}{5}$

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