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

Apr 4, 2016

$T = - \frac{7}{2} + \frac{\sqrt{57}}{2}$ or $T = - \frac{7}{2} - \frac{\sqrt{57}}{2}$

#### Explanation:

As ${\left(x + a\right)}^{2} = {x}^{2} + 2 a x + {a}^{2}$, to complete let us say

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

Hence to solve ${T}^{2} + 7 T - 2 = 0$ we should add and subtract ${\left(\frac{7}{2}\right)}^{2} = \frac{49}{4}$.

Hence, ${T}^{2} + 7 T - 2 = 0$ can be written as ${T}^{2} + 7 T + \frac{49}{4} - \frac{49}{4} - 2 = 0$ or

${T}^{2} + 7 T + \frac{49}{4} = \frac{49}{4} + 2 = \frac{57}{4}$ or

= ${\left(T + \frac{7}{2}\right)}^{2} = \frac{57}{4}$

or $T + \frac{7}{2} = \frac{\sqrt{57}}{2}$ or $T + \frac{7}{2} = - \frac{\sqrt{57}}{2}$

or $T = - \frac{7}{2} + \frac{\sqrt{57}}{2}$ or $T = - \frac{7}{2} - \frac{\sqrt{57}}{2}$