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how would I solve the equation, 3y^2-4y-15=0
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quadratically .... either complete the square or use the formula
ax^2 + bx+c = 0 \[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
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youre welcome
You can also try factoring.
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\(3y^2 - 4y - 15 = 0\) \(3y^2 - 9y + 5y - 15 = 0\) \(3y(y - 3) + 5(y - 3) = 0\) \( (y - 3)(3y + 5) = 0\) Now set each factor equal to zero and solve for y.
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