Question 74404
{{{ 5x^2 + 8x = -3 }}}
First you must get everything on the left side thereby setting the equation equal to zero.
{{{ 5x^2 + 8x + 3 = 0 }}}
This is now in the proper form.
A = 5
B = 8
C = 3

Plug these values into the Quadratic formula
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{ x = (-8 +- sqrt(8^2-4*5*3))/(2*5) }}}
square 8
{{{ x = (-8 +- sqrt(64-4*5*3))/(2*5) }}}
Multiply 4, 5 and 3
{{{ x = (-8 +- sqrt(64-60))/(2*5) }}}
Multiply 2 and 5
{{{ x = (-8 +- sqrt(64-60))/10 }}}
Subtract 64 and 60
{{{ x = (-8 +- sqrt(4))/10 }}}
this is where it breaks into TWO problems
{{{ x = (-8 + sqrt(4))/10 }}} and {{{ x = (-8 - sqrt(4))/10 }}}
Square root 4 in both problems
{{{ x = (-8 + 2)/10 }}} and {{{ x = (-8 - 2)/10 }}}
work the Numerator in each fraction
{{{ x = (-6)/10 }}} and {{{ x = (-10)/10 }}}
Reduce the fractions
{{{ x = -3/5 }}} and {{{ x = -1 }}}