How do you solve $T^2 + 7T - 2 = 0$ by completing the square?

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Answer 1 · 2016-04-04T10:35:33

$T=-7/2+sqrt57/2$ or $T=-7/2-sqrt57/2$

As $(x+a)^2=x^2+2ax+a^2$ , to complete let us say

$x^2+2ax$ to make complete square, we should add $a^2$ or square of half of the coefficient of x.

Hence to solve $T^2+7T-2=0$ we should add and subtract $(7/2)^2=49/4$ .

Hence, $T^2+7T-2=0$ can be written as $T^2+7T+49/4-49/4-2=0$ or

$T^2+7T+49/4=49/4+2=57/4$ or

= $(T+7/2)^2=57/4$

or $T+7/2=sqrt57/2$ or $T+7/2=-sqrt57/2$

or $T=-7/2+sqrt57/2$ or $T=-7/2-sqrt57/2$

How do you solve T^2 + 7T - 2 = 0T^2 + 7T - 2 = 0 by completing the square?