Question

[(2x+3)+(x+5)]^2+[(2x+3)-(x+5)]^2=10x^2+92

Answer 1

${((2x + 3) + (x + 5))}^{2} + {((2x + 3) - (x + 5))}^{2} = 10 {x}^{2} + 92$
${(2x + 3)}^{2} + {(x + 5)}^{2} + 2(2x + 3)(x + 5) + {(2x + 3)}^{2} + {(x + 5)}^{2} - 2(2x + 3)(x + 5) = 10 {x}^{2} + 92$

$2 {(2x + 3)}^{2} + 2 {(x + 5)}^{2} = 10 {x}^{2} + 92$
$2(4 {x}^{2} + 9 + 12x) + 2( {x}^{2} + 25 + 10x) = 10 {x}^{2} + 92$
$2 {x}^{2} + 18 + 24x + 2 {x}^{2} + 50 + 20x = 10 {x}^{2} + 92$
$4 {x}^{2} + 44x + 68 = 10 {x}^{2} + 92$
$6 {x}^{2} - 44x + 24 = 0$
$3 {x}^{2} - 22x + 12 = 0$
$x = \frac{11 + \sqrt{85} }{3} or \frac{11 - \sqrt{85} }{3}$