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5x^{2}-4x=3
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.
5x^{2}-4x-3=3-3
Subtract 3 from both sides of the equation.
5x^{2}-4x-3=0
Subtracting 3 from itself leaves 0.
x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 5\left(-3\right)}}{2\times 5}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -4 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-4\right)±\sqrt{16-4\times 5\left(-3\right)}}{2\times 5}
Square -4.
x=\frac{-\left(-4\right)±\sqrt{16-20\left(-3\right)}}{2\times 5}
Multiply -4 times 5.
x=\frac{-\left(-4\right)±\sqrt{16+60}}{2\times 5}
Multiply -20 times -3.
x=\frac{-\left(-4\right)±\sqrt{76}}{2\times 5}
Add 16 to 60.
x=\frac{-\left(-4\right)±2\sqrt{19}}{2\times 5}
Take the square root of 76.
x=\frac{4±2\sqrt{19}}{2\times 5}
The opposite of -4 is 4.
x=\frac{4±2\sqrt{19}}{10}
Multiply 2 times 5.
x=\frac{2\sqrt{19}+4}{10}
Now solve the equation x=\frac{4±2\sqrt{19}}{10} when ± is plus. Add 4 to 2\sqrt{19}.
x=\frac{\sqrt{19}+2}{5}
Divide 4+2\sqrt{19} by 10.
x=\frac{4-2\sqrt{19}}{10}
Now solve the equation x=\frac{4±2\sqrt{19}}{10} when ± is minus. Subtract 2\sqrt{19} from 4.
x=\frac{2-\sqrt{19}}{5}
Divide 4-2\sqrt{19} by 10.
x=\frac{\sqrt{19}+2}{5} x=\frac{2-\sqrt{19}}{5}
The equation is now solved.
5x^{2}-4x=3
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{5x^{2}-4x}{5}=\frac{3}{5}
Divide both sides by 5.
x^{2}-\frac{4}{5}x=\frac{3}{5}
Dividing by 5 undoes the multiplication by 5.
x^{2}-\frac{4}{5}x+\left(-\frac{2}{5}\right)^{2}=\frac{3}{5}+\left(-\frac{2}{5}\right)^{2}
Divide -\frac{4}{5}, the coefficient of the x term, by 2 to get -\frac{2}{5}. Then add the square of -\frac{2}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{4}{5}x+\frac{4}{25}=\frac{3}{5}+\frac{4}{25}
Square -\frac{2}{5} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{4}{5}x+\frac{4}{25}=\frac{19}{25}
Add \frac{3}{5} to \frac{4}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{2}{5}\right)^{2}=\frac{19}{25}
Factor x^{2}-\frac{4}{5}x+\frac{4}{25}. 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{2}{5}\right)^{2}}=\sqrt{\frac{19}{25}}
Take the square root of both sides of the equation.
x-\frac{2}{5}=\frac{\sqrt{19}}{5} x-\frac{2}{5}=-\frac{\sqrt{19}}{5}
Simplify.
x=\frac{\sqrt{19}+2}{5} x=\frac{2-\sqrt{19}}{5}
Add \frac{2}{5} to both sides of the equation.