How do you use the quadratic formula to solve #5s^2+6x+3=0#?

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Answer 1 · 2018-01-30T01:14:39

#x = 0.2*(-3 + sqrt6color(red) i), (0.2 * (-3 - sqrt6color(red) i)#

#x = 0.2 * (-3 +- 2.4495 color(red)i)#

Given Equation : #5x^2 + 6x + 3 = 0#

Formula for finding the routes : #(-b +- sqrt(b^2 - (4 a c))) / (2a)#

where a, b c are the coefficients of #x^2, x and const# terms respy.

# x = (-6 +- sqrt(6^2 - (4* 5 * 3))) / (2*5)#

#x = (-6 +- sqrt(-24))/10#

# x = -0.6 +- 0.2 sqrt6 i#

#x = -0.6 + 0.2 sqrt6color(red) i, -0.6 - 0.2 sqrt6 color(red)i#