Question 247449
{{{6y + 2 = 2y^2 - 6}}}
Subtract 2 from both sides
{{{6y = 2y^2 - 8}}}
Subtract 6y from both sides
{{{0 = 2y^2 - 6y - 8}}}
Divide both sides by 2
{{{0 = y^2 - 3y - 4}}}
Which means
{{{y^2 - 3y - 4 = 0}}}
Can you factor -4 to find two factors that total -3?
How about -4 and + 1?
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{{{(y-4)(y+1)=0}}}
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So, y = 4 or -1.
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Always check your work!
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Substitute y =4.
{{{6y + 2 = 2y^2 - 6}}}
{{{6(4) + 2 = 2(4^2) - 6}}}
{{{24 + 2 = 2(16) - 6}}}
{{{26 = 26}}}
So y =4 checks.
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Substitute y=-1.
{{{6(-1) + 2 = 2(-1^2) - 6}}}
{{{-6 + 2 = 2(1) - 6}}}
{{{-4 = -4}}}
So y = -1 checks.
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Done.