Question 507830
You're OK so far:
{{{5y^2+2y = 24}}} Subtract 24 from both sides.
{{{5y^2+2y-24 = 0}}} Now you have a quadratic equation which you can solve using the quadratic formula: {{{y = (-b+-sqrt(b^2-4ac))/2a}}} and here...a=5, b=2, and c=-24. Make the appropriate substitutions:
{{{y = (-2+-sqrt(2^2-4(5)(-24)))/2(5)}}} Evaluate:
{{{y = (-2+-sqrt(4-(-480)))/10}}}
{{{y = (-2+-sqrt(484))/10}}}
{{{y[1] = -0.2+2.2}}} or {{{y[2] = -0.2-2.2}}}
{{{y[1] = 2}}} or {{{y[2] = -2.4}}}
{{{x[1] = 2+5y}}}
{{{x[1] = 2+5(2)}}}
{{{x[1] = 12}}} or
{{{x[2] = 2+5(-2.4)}}}
{{{x[2] = 2+(-12)}}}
{{{x[2] = -10}}}
So there are two pairs of numbers that will satisfy the given constraints:
2 and 12
-2.4 and -10