Question

solve this equation :
${x}^{2} + 11x + 28 = 0$
​

Answer 1

x=-4, -7

Step-by-step explanation:

${x}^{2} + 11x + 28 = 0 \\ {x}^{2} + (7 + 4)x + 28 = 0 \\ {x}^{2} + 7x + 4x + 28 = 0 \\ {x}(x + 7)+4(x + 7) = 0 \\ (x + 7)(x + 4) = 0\\ if \\ (x + 7) = 0 \: \: x = - 7 \\ similarly\: \: \\ (x + 4) = 0 \: \: x = - 4$

I hope this help you

Answer 2

Answer:

${x}^{2} + 11x + 28 = 0$

$= {x}^{2} + 7x + 4x + 28 = 0$

$= {x}^{2} + 7x + 4x + 28 = 0$

$= x(x+ 7) + 4(x+ 7) = 0$

$= (x + 7) (x + 4) = 0$

$(x + 7) = 0. (x + 4) = 0$

$x = -7$

$x = - 4$

Hope it helps you..✌️