How do you solve #4h ^ { 2} - 6h + 7= 0#?

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Answer 1 · 2017-07-20T21:56:46

See a solution process below.

We can use the quadratic equation to solve this problem:

For #ax^2 + bx + c = 0#, the values of #x# which are the solutions to the equation are given by:

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

Substituting:

#color(red)(4)# for #color(red)(a)#

#color(blue)(-6)# for #color(blue)(b)#

#color(green)(7)# for #color(green)(c)# gives:

#x = (-color(blue)((-6)) +- sqrt(color(blue)((-6))^2 - (4 * color(red)(4) * color(green)(7))))/(2 * color(red)(4))#

#x = (color(blue)(6) +- sqrt(36 - 112))/8#

#x = (color(blue)(6) +- sqrt(-76))/8#

#x = (color(blue)(6) +- sqrt(4 * -19))/8#

#x = (color(blue)(6) +- sqrt(4)sqrt(-19))/8#

#x = (color(blue)(6) +- 2sqrt(-19))/8#

Or

#x = color(blue)(6)/8 + (2sqrt(-19))/8# and #x = color(blue)(6)/8 - (2sqrt(-19))/8#

#x = 3/4 + sqrt(-19)/4# and #x = 3/4 - sqrt(-19)/4#