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x\left(0.9x-3\right)=0
Factor out x.
x=0 x=\frac{10}{3}
To find equation solutions, solve x=0 and \frac{9x}{10}-3=0.
0.9x^{2}-3x=0
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.
x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}}}{2\times 0.9}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 0.9 for a, -3 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-3\right)±3}{2\times 0.9}
Take the square root of \left(-3\right)^{2}.
x=\frac{3±3}{2\times 0.9}
The opposite of -3 is 3.
x=\frac{3±3}{1.8}
Multiply 2 times 0.9.
x=\frac{6}{1.8}
Now solve the equation x=\frac{3±3}{1.8} when ± is plus. Add 3 to 3.
x=\frac{10}{3}
Divide 6 by 1.8 by multiplying 6 by the reciprocal of 1.8.
x=\frac{0}{1.8}
Now solve the equation x=\frac{3±3}{1.8} when ± is minus. Subtract 3 from 3.
x=0
Divide 0 by 1.8 by multiplying 0 by the reciprocal of 1.8.
x=\frac{10}{3} x=0
The equation is now solved.
0.9x^{2}-3x=0
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{0.9x^{2}-3x}{0.9}=\frac{0}{0.9}
Divide both sides of the equation by 0.9, which is the same as multiplying both sides by the reciprocal of the fraction.
x^{2}+\left(-\frac{3}{0.9}\right)x=\frac{0}{0.9}
Dividing by 0.9 undoes the multiplication by 0.9.
x^{2}-\frac{10}{3}x=\frac{0}{0.9}
Divide -3 by 0.9 by multiplying -3 by the reciprocal of 0.9.
x^{2}-\frac{10}{3}x=0
Divide 0 by 0.9 by multiplying 0 by the reciprocal of 0.9.
x^{2}-\frac{10}{3}x+\left(-\frac{5}{3}\right)^{2}=\left(-\frac{5}{3}\right)^{2}
Divide -\frac{10}{3}, the coefficient of the x term, by 2 to get -\frac{5}{3}. Then add the square of -\frac{5}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{10}{3}x+\frac{25}{9}=\frac{25}{9}
Square -\frac{5}{3} by squaring both the numerator and the denominator of the fraction.
\left(x-\frac{5}{3}\right)^{2}=\frac{25}{9}
Factor x^{2}-\frac{10}{3}x+\frac{25}{9}. 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-\frac{5}{3}\right)^{2}}=\sqrt{\frac{25}{9}}
Take the square root of both sides of the equation.
x-\frac{5}{3}=\frac{5}{3} x-\frac{5}{3}=-\frac{5}{3}
Simplify.
x=\frac{10}{3} x=0
Add \frac{5}{3} to both sides of the equation.