Question

Реши уравнение:
x^2+17x+52=0

Answer 1

$x {}^{2} + 17x + 52 = 0$

${x}^{2} + 13x + 4x + 52 = 0$

$x times (x + 13) + 4(x + 13) = 0$

$(x + 13) times (x + 4) = 0$

$x + 13 = 0$

$x + 4 = 0$

$x = - 13$

$x = - 4$
$x1 = - 13$
$\ \ x2 = - 4$


Удачи!

Answer 2

${x}^{2} + 17x + 52 = 0 \ {x}^{2} + 13x + 4x + 52 = 0 \ x times (x + 13) + 4 times (x - + 13) = 0 \ (x + 13) times (x + 4) = 0 \ x + 13 = 0 \ x + 4 = 0 \ x = - 13 \ x = - 4 \ x1 = - 13 \ x2 = - 4$