How do you solve #w+ 5= - 6w ^ { 2}#?

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Answer 1 · 2017-09-15T13:27:42

See a solution process below:

First, add #color(red)(6w^2)# to each side of the equation to put this equation into standard form while keeping the equation balanced:

#color(red)(6w^2) + w + 5 = color(red)(6w^2) - 6w^2#

#6w^2 + w + 5 = 0#

We can then use the quadratic equation to solve this problem:

For #color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0#, the values of #x# which are the solutions to the equation are given by:

#x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))#

Substituting:

#color(red)(6)# for #color(red)(a)#

#color(blue)(1)# for #color(blue)(b)#

#color(green)(5)# for #color(green)(c)# gives:

#x = (-color(blue)(1) +- sqrt(color(blue)(1)^2 - (4 * color(red)(6) * color(green)(5))))/(2 * color(red)(6))#

#x = (-color(blue)(1) +- sqrt(1 - 120))/12#

#x = (-color(blue)(1) +- sqrt(-119))/12#