Question 813463: Solve using the Quadratic Formumla:
2y^2-(y+2)(y-3)=12
Please and Thanks!
Answer by TimothyLamb(4379)   (Show Source):
You can put this solution on YOUR website! 2y^2 - (y + 2)(y - 3) = 12
2y^2 - (y^2 - y - 6) = 12
2y^2 - y^2 + y + 6 = 12
y^2 + y - 6 = 0
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the above quadratic equation is in standard form, with a=1, b=1, and c=-6
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to solve the quadratic equation, plug this:
1 1 -6
into this: https://sooeet.com/math/quadratic-equation-solver.php
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Answer 1:
the vertex of the quadratic equation is a minimum point at: ( -0.5, -6.25 )
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Answer 2:
the quadratic equation has two real roots:
root1 = 2
root2 = -3
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