Question 60923
You are solving a quadratic equation here.

{{{4x^2+5x-6 = 0}}}

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} with {{{a=4}}} {{{b=5}}} {{{c=-6}}}

Then, {{{x = (-5 +- sqrt(5^2-4(4)(-6)))/(2*4)}}}
Or, {{{x = (-5 +- sqrt(25+16*6))/8}}}
Or, {{{x = (-5 +- sqrt(121))/8}}}
Or, {{{x = (-5 +- 11)/8}}}
Or, {{{x = -16/8 = -2}}} and {{{x = 6/8 = 3/4}}}

Verify these values in the original equation:

{{{4((-2)^2)+5(-2)-6 = 4*4-10-6}}} which is indeed 0.
{{{4(3/4)^2+5(3/4)-6 = 4(9/16)+15/4-6 = (36/16)+(15/4)-6 = (9/4)+(15/4)-6 = 0}}}