# How do you solve 10t^2 - 29t = -10?

Jul 28, 2018

$t = \frac{2}{5} \mathmr{and} \frac{5}{2}$

#### Explanation:

$10 {t}^{2} - 29 t + 10 = 0$
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,
$10 {t}^{2} - 25 t - 4 t + 10 = 0$
$5 t \left(2 t - 5\right) - 2 \left(2 t - 5\right) = 0$
$\left(5 t - 2\right) \left(2 t - 5\right) = 0$
which implies that,
$t = \frac{2}{5} \mathmr{and} \frac{5}{2}$

Jul 28, 2018

$t = \frac{5}{2} \mathmr{and} t = \frac{2}{5}$

#### Explanation:

Here,

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

$10 {t}^{2} - 29 t + 10 = 0$

$10 {t}^{2} - 25 t - 4 t + 10 = 0$

$5 t \left(2 t - 5\right) - 2 \left(2 t - 5\right) = 0$

$\left(2 t - 5\right) \left(5 t - 2\right) = 0$

$2 t - 5 = 0 \mathmr{and} 5 t - 2 = 0$

$2 t = 5 \mathmr{and} 5 t = 2$

$t = \frac{5}{2} \mathmr{and} t = \frac{2}{5}$

...........................................OR......................................

$10 {t}^{2} - 29 t + 10 = 0$

Comparing with $a {t}^{2} + b t + c = 0$

$a = 10 , b = - 29 \mathmr{and} c = 10$

So,

$t = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$\therefore t = \frac{29 \pm \sqrt{{29}^{2} - 4 \left(10\right) \left(10\right)}}{2 \times 10}$

$\therefore t = \frac{29 \pm \sqrt{841 - 400}}{20}$

$\therefore t = \frac{29 \pm 21}{20}$

$\therefore t = \frac{29 + 21}{20} \mathmr{and} t = \frac{29 - 21}{20}$

$\therefore t = \frac{50}{20} \mathmr{and} t = \frac{8}{20}$

$\therefore t = \frac{5}{2} \mathmr{and} t = \frac{2}{5}$