Question

Реши уравнение:
11t2+44t−(t+4)=0.

Корни уравнения
t1=
;t2=
.

Answer 1

$t_1=1/11\\t_2=-4$

Объяснение:

$11t^2+44t-(t+4)=0\\11t(t+4)-(t+4)=0\\(11t-1)(t+4)=0\\t=1/11\\t=-4$

Answer 2

t1 = 1/11 ; t2 = -4

Объяснение:

$11t^{2} + 44t - (t + 4) = 0$

$11t^{2} + 44t - t - 4 = 0$

$11t^{2} + 43t - 4 = 0$

$D = b^{2} - 4ac$

$D = 43^{2} - 4*11*(-4)$

$D = 1849 - (-176)$

$D = 1849 + 176$

$\sqrt{D} = \sqrt{2025} = 45$

$D = 2025$

$x_{1} = \frac{-b + \sqrt{D} }{2a} = \frac{-43 + 45}{22} = \frac{2}{22} = \frac{1}{11}$

$x_{2} = \frac{-b - \sqrt{D} }{2a} = \frac{-43 - 45}{22} = \frac{-88}{22} = -4$