[tex] {x}^{2} = 3x + 10 \\ {x}^{2} - 3x - 10 = 0 \\ {(x)}^{2} - 2(x)( \frac{3}{2} ) + {( \frac{3}{2} )}^{2} - \frac{49}{4} = 0 \\ {(x - \frac{3}{2} )}^{2} - {( \frac{7}{2} )}^{2} = 0 \\ {(x - \frac{3}{2} } - \frac{7}{2} ) {(x - \frac{3}{2} } + \frac{7}{2} ) = 0 \\ {(x - \frac{10}{2} } ) {(x } + \frac{4}{2} ) = 0 \\ (x - 5)(x + 2) = 0 \\ then \: x = 5 \: \: \: or \: \: \: x = - 2[/tex]
but x cannot be negative so x= 5 is the answer
