Question

factorise x^2+(x+1)^2=613

Answer 1

$\underline{\large\bf{\mathfrak{Hello!}}}$

$= > {x}^{2} + {(x + 1)}^{2} = 613 \\ = > {x}^{2} + {x}^{2} + 2x + 1 = 613 \\ = > 2 {x}^{2} + 2 x + 1 - 613 = 0 \\ = > 2 {x}^{2} + 2x - 612 = 0 \\ = > 2( {x}^{2} + 2 - 306) = 0 \\ = > {x}^{2} + 2 - 306 = 0 \\ = > {x}^{2} + 18x - 17x - 306 = 0 \\ = > x(x + 18) - 17(x + 18) = 0 \\ = > (x + 18)(x - 17) = 0 \\ \\ = > x = - 18 \\ \: \: \: \: \: \: \: \: x = 17$

$\boxed{x = -18 \: or \: 17}$

$\bf{\mathfrak{Hope \: this \: helps...:)}}$

Answer 2

$\huge\underline\mathfrak\blue{Solution}$

x² + (x + 1)² = 613

x² + x² + 1 + 2x = 613

2x² + 2x - 612 = 0

x² + x - 306 = 0

x² + 18x - 17x - 306 = 0

x (x + 18) - 17 (x + 18) = 0

Hence,

x = 17 or x = -18

Cheers!! ✨✨