Solve 2x2 + 20x = - 38.

a. - 5 ± square root 14

b.-5 ± square root 51

c. - 5 ±square root 6

d.-5 ± square root 13

2x^2+20x=-38 divide both sides by 2

x^2+10x=-19, halve the linear coefficient, square it, than add that value to both sides of the equation, in this case add (10/2) ^2=25 ...

x^2+10x+25=6 now the left side is a perfect square

(x+5) ^2=6 take the square root of both sides

x+5=±√6 subtract 5 from both sides

x=-5±√6 (so answer c.)