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

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