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Mathematics 15 Online
OpenStudy (anonymous):

how would I solve the equation, 3y^2-4y-15=0

OpenStudy (amistre64):

quadratically .... either complete the square or use the formula

OpenStudy (amistre64):

ax^2 + bx+c = 0 \[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

OpenStudy (anonymous):

thank you

OpenStudy (amistre64):

youre welcome

OpenStudy (mathstudent55):

You can also try factoring.

OpenStudy (mathstudent55):

\(3y^2 - 4y - 15 = 0\) \(3y^2 - 9y + 5y - 15 = 0\) \(3y(y - 3) + 5(y - 3) = 0\) \( (y - 3)(3y + 5) = 0\) Now set each factor equal to zero and solve for y.

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