Question 574851
Solve the quadratic equation 2x^2-3x-2=0 in quadratic "form" .
 How so i solve this problem. 
If you can please show me step from step.
:
2x^2 - 3x - 2 = 0; this form ax^2 + bx + c = 0; therefore
a = 2; b = -3; c = -2
the quadratic formula:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
Replace a, b, c
{{{x = (-(-3) +- sqrt(-3^2-4*2*-2 ))/(2*2) }}}
Do the math, minus a minus is a plus
{{{x = (3 +- sqrt(9-(-16) ))/4 }}}
;
{{{x = (3 +- sqrt(9+16 ))/4 }}}
:
{{{x = (3 +- sqrt(25 ))/4 }}}
Find the square root, we have two solutions
{{{x = (3 + 5 ))/4 }}}
x = {{{8/4}}}
x = 2
and
{{{x = (3 - 5 ))/4 }}}
x = {{{(-2)/4}}}
x = -.5
:
:
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