Solve for y
y=\frac{x\left(3x-40\right)}{12}
Solve for x (complex solution)
x=\frac{2\sqrt{9y+100}+20}{3}
x=\frac{-2\sqrt{9y+100}+20}{3}
Solve for x
x=\frac{2\sqrt{9y+100}+20}{3}
x=\frac{-2\sqrt{9y+100}+20}{3}\text{, }y\geq -\frac{100}{9}
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-40x-12y=-3x^{2}
Subtract 3x^{2} from both sides. Anything subtracted from zero gives its negation.
-12y=-3x^{2}+40x
Add 40x to both sides.
-12y=40x-3x^{2}
The equation is in standard form.
\frac{-12y}{-12}=\frac{x\left(40-3x\right)}{-12}
Divide both sides by -12.
y=\frac{x\left(40-3x\right)}{-12}
Dividing by -12 undoes the multiplication by -12.
y=\frac{x^{2}}{4}-\frac{10x}{3}
Divide x\left(40-3x\right) by -12.
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