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9x+3-11x=25x^{2}+1\times 3
Use the distributive property to multiply 3 by 3x+1.
-2x+3=25x^{2}+1\times 3
Combine 9x and -11x to get -2x.
-2x+3=25x^{2}+3
Multiply 1 and 3 to get 3.
-2x+3-25x^{2}=3
Subtract 25x^{2} from both sides.
-2x+3-25x^{2}-3=0
Subtract 3 from both sides.
-2x-25x^{2}=0
Subtract 3 from 3 to get 0.
x\left(-2-25x\right)=0
Factor out x.
x=0 x=-\frac{2}{25}
To find equation solutions, solve x=0 and -2-25x=0.
9x+3-11x=25x^{2}+1\times 3
Use the distributive property to multiply 3 by 3x+1.
-2x+3=25x^{2}+1\times 3
Combine 9x and -11x to get -2x.
-2x+3=25x^{2}+3
Multiply 1 and 3 to get 3.
-2x+3-25x^{2}=3
Subtract 25x^{2} from both sides.
-2x+3-25x^{2}-3=0
Subtract 3 from both sides.
-2x-25x^{2}=0
Subtract 3 from 3 to get 0.
-25x^{2}-2x=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(-2\right)±\sqrt{\left(-2\right)^{2}}}{2\left(-25\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -25 for a, -2 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-2\right)±2}{2\left(-25\right)}
Take the square root of \left(-2\right)^{2}.
x=\frac{2±2}{2\left(-25\right)}
The opposite of -2 is 2.
x=\frac{2±2}{-50}
Multiply 2 times -25.
x=\frac{4}{-50}
Now solve the equation x=\frac{2±2}{-50} when ± is plus. Add 2 to 2.
x=-\frac{2}{25}
Reduce the fraction \frac{4}{-50} to lowest terms by extracting and canceling out 2.
x=\frac{0}{-50}
Now solve the equation x=\frac{2±2}{-50} when ± is minus. Subtract 2 from 2.
x=0
Divide 0 by -50.
x=-\frac{2}{25} x=0
The equation is now solved.
9x+3-11x=25x^{2}+1\times 3
Use the distributive property to multiply 3 by 3x+1.
-2x+3=25x^{2}+1\times 3
Combine 9x and -11x to get -2x.
-2x+3=25x^{2}+3
Multiply 1 and 3 to get 3.
-2x+3-25x^{2}=3
Subtract 25x^{2} from both sides.
-2x-25x^{2}=3-3
Subtract 3 from both sides.
-2x-25x^{2}=0
Subtract 3 from 3 to get 0.
-25x^{2}-2x=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{-25x^{2}-2x}{-25}=\frac{0}{-25}
Divide both sides by -25.
x^{2}+\left(-\frac{2}{-25}\right)x=\frac{0}{-25}
Dividing by -25 undoes the multiplication by -25.
x^{2}+\frac{2}{25}x=\frac{0}{-25}
Divide -2 by -25.
x^{2}+\frac{2}{25}x=0
Divide 0 by -25.
x^{2}+\frac{2}{25}x+\left(\frac{1}{25}\right)^{2}=\left(\frac{1}{25}\right)^{2}
Divide \frac{2}{25}, the coefficient of the x term, by 2 to get \frac{1}{25}. Then add the square of \frac{1}{25} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{2}{25}x+\frac{1}{625}=\frac{1}{625}
Square \frac{1}{25} by squaring both the numerator and the denominator of the fraction.
\left(x+\frac{1}{25}\right)^{2}=\frac{1}{625}
Factor x^{2}+\frac{2}{25}x+\frac{1}{625}. 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{1}{25}\right)^{2}}=\sqrt{\frac{1}{625}}
Take the square root of both sides of the equation.
x+\frac{1}{25}=\frac{1}{25} x+\frac{1}{25}=-\frac{1}{25}
Simplify.
x=0 x=-\frac{2}{25}
Subtract \frac{1}{25} from both sides of the equation.