## Algebra 1

$t=4$
$2t = \sqrt {2t +56}$ $(2t)^2 = (\sqrt {2t +56})^2$ $4t^2 = 2t+56$ $4t^2-2t-56 = 2t+56-2t-56$ $4t^2-2t-56 =0$ $(4t^2-2t-56 =0)/2$ $2t^2-t-28 =0$ $(2t+7)(t-4) = 0$ $2t+7=0$ $2t=-7$ $2t/2 = -7/2$ $t = -3.5$ $t-4=0$ $t=4$ $t = -3.5$ $2t = \sqrt {2t +56}$ $2*-3.5 = \sqrt {2*-3.5 +56}$ $-7 = \sqrt {-7 + 56}$ (we can't have a negative number as the square root of a number, so this answer is invalid)