Solve 2x2 + 20x = −38. a) −5 ± square root 14 b) −5 ± square root 51 c) −5 ±square root 6 d) −5 ± square root 13

Correct Answer is C(−5 ±square root 6)

Step-by-step explanation:

2x^2+20x+38=0

2(x^2+10x+19)=0

x^2+10x+19=0

This is a quadratic equation here we have used following formula to find the roots of this equation

x=-b−±square root (b^2-4*a*c))/2*a(1)

Here,

a=1

b=10

c=19

put values of a, b and c in equation (1)

x=(-10 ±square root(10^2-4*1*19))/2*1

x=(-10 ±square root(100-76))/2

x=(-10 ±square root(24))/2

x=(-10 ±square root(2*2*6))/2

x=(-10 ±2square root(6))/2

x=2(-5 ±square root(6))/2

x=-5 ±square root(6)

Hope this answer will help you

answer: the answer to your question is parallel