Question

(t+3)2+(t+4)2=(2t+5)2-2(t2-3)

Answer 1

Tu ejercicio.

$(t+3)2+(t+4)2=(2t+5)2-2(t2-3)$

Respuesta:

$t_1 = 1\\\\t_2 = -1$

Explicación paso a paso:

$(2t + 6) + (2t + 8) - {(4t + 10) - (2t ^ 2-6)} = 0\\\\(4t + 14) - {- 2t ^ 2 + 4t + 16} = 0\\\\2t ^ 2-2 = 0\\\\(2) * t ^ 2 + (- 2) = 0\\\\t = (+ - raiz cuadrada (- (2) * (- 2))) / (2)\\\\t = (+ - raiz cuadrada (4)) / (2)\\\\t = (+ - (2)) / (2)\\\\t = + - (\ cancelar {(2)}) / (\ cancelar {(2)})\\\\t_1 = 1\\\\t_2 = -1$