If $a = \frac{b^{2} - c}{2}$ $a=(b^2-c)/2$ , how do you find the value of a when b=2, c=-4?

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Answer 1 · 2017-01-12T09:45:59

$a = 4$ $a=4$

WE have $b = 2$ $b=2$ hence $b^{2} = 2 \times 2 = 4$ $b^2=2xx2=4$ and have $c = - 4$ $c=-4$

Hence $a = \frac{b^{2} - c}{2} = \frac{4 - (- 4)}{2} = \frac{4 + 4}{2} = \frac{8}{2} = 4$ $a=(b^2-c)/2=(4-(-4))/2=(4+4)/2=8/2=4$

If a=b2−c2a=(b^2-c)/2, how do you find the value of a when b=2, c=-4?