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.

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

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

$x = -0.6 +- 0.2 sqrt6 i$

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

How do you use the quadratic formula to solve 5s^2+6x+3=05s^2+6x+3=0?