How do you solve #y = 3x^2 + 5x# using the quadratic formula?

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Answer 1 · 2018-06-17T00:27:33

#x=0,##-5/3#

#y=3x^2+5x# is a quadratic equation in standard form:

#y=ax^2+bx+c#,

where:

#a=3#, #b=5#, #c=0#

To solve the equation for #x#, substitute #0# for #y#.

#0=3x^2+5x#

#x=(-b+-sqrt(b^2-4ac))/(2a)#

Plug in the known values and solve.

#x=(-5+-sqrt(5^2-4*3*0))/(2*3)#

#x=(-5+-sqrt25)/6#

#x=(-5+-5)/6#

#x=(-5+5)/6,# #(-5-5)/6#

#x=0/6, -10/6#

Simplify.

#x=0,##-5/3#