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-6x^{2}+72x=216
All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.
-6x^{2}+72x-216=216-216
Subtract 216 from both sides of the equation.
-6x^{2}+72x-216=0
Subtracting 216 from itself leaves 0.
x=\frac{-72±\sqrt{72^{2}-4\left(-6\right)\left(-216\right)}}{2\left(-6\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -6 for a, 72 for b, and -216 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-72±\sqrt{5184-4\left(-6\right)\left(-216\right)}}{2\left(-6\right)}
Square 72.
x=\frac{-72±\sqrt{5184+24\left(-216\right)}}{2\left(-6\right)}
Multiply -4 times -6.
x=\frac{-72±\sqrt{5184-5184}}{2\left(-6\right)}
Multiply 24 times -216.
x=\frac{-72±\sqrt{0}}{2\left(-6\right)}
Add 5184 to -5184.
x=-\frac{72}{2\left(-6\right)}
Take the square root of 0.
x=-\frac{72}{-12}
Multiply 2 times -6.
x=6
Divide -72 by -12.
-6x^{2}+72x=216
Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.
\frac{-6x^{2}+72x}{-6}=\frac{216}{-6}
Divide both sides by -6.
x^{2}+\frac{72}{-6}x=\frac{216}{-6}
Dividing by -6 undoes the multiplication by -6.
x^{2}-12x=\frac{216}{-6}
Divide 72 by -6.
x^{2}-12x=-36
Divide 216 by -6.
x^{2}-12x+\left(-6\right)^{2}=-36+\left(-6\right)^{2}
Divide -12, the coefficient of the x term, by 2 to get -6. Then add the square of -6 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-12x+36=-36+36
Square -6.
x^{2}-12x+36=0
Add -36 to 36.
\left(x-6\right)^{2}=0
Factor x^{2}-12x+36. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.
\sqrt{\left(x-6\right)^{2}}=\sqrt{0}
Take the square root of both sides of the equation.
x-6=0 x-6=0
Simplify.
x=6 x=6
Add 6 to both sides of the equation.
x=6
The equation is now solved. Solutions are the same.