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3x^{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.
3x^{2}-4x-3=3-3
Subtract 3 from both sides of the equation.
3x^{2}-4x-3=0
Subtracting 3 from itself leaves 0.
x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 3\left(-3\right)}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 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 3\left(-3\right)}}{2\times 3}
Square -4.
x=\frac{-\left(-4\right)±\sqrt{16-12\left(-3\right)}}{2\times 3}
Multiply -4 times 3.
x=\frac{-\left(-4\right)±\sqrt{16+36}}{2\times 3}
Multiply -12 times -3.
x=\frac{-\left(-4\right)±\sqrt{52}}{2\times 3}
Add 16 to 36.
x=\frac{-\left(-4\right)±2\sqrt{13}}{2\times 3}
Take the square root of 52.
x=\frac{4±2\sqrt{13}}{2\times 3}
The opposite of -4 is 4.
x=\frac{4±2\sqrt{13}}{6}
Multiply 2 times 3.
x=\frac{2\sqrt{13}+4}{6}
Now solve the equation x=\frac{4±2\sqrt{13}}{6} when ± is plus. Add 4 to 2\sqrt{13}.
x=\frac{\sqrt{13}+2}{3}
Divide 4+2\sqrt{13} by 6.
x=\frac{4-2\sqrt{13}}{6}
Now solve the equation x=\frac{4±2\sqrt{13}}{6} when ± is minus. Subtract 2\sqrt{13} from 4.
x=\frac{2-\sqrt{13}}{3}
Divide 4-2\sqrt{13} by 6.
x=\frac{\sqrt{13}+2}{3} x=\frac{2-\sqrt{13}}{3}
The equation is now solved.
3x^{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{3x^{2}-4x}{3}=\frac{3}{3}
Divide both sides by 3.
x^{2}-\frac{4}{3}x=\frac{3}{3}
Dividing by 3 undoes the multiplication by 3.
x^{2}-\frac{4}{3}x=1
Divide 3 by 3.
x^{2}-\frac{4}{3}x+\left(-\frac{2}{3}\right)^{2}=1+\left(-\frac{2}{3}\right)^{2}
Divide -\frac{4}{3}, the coefficient of the x term, by 2 to get -\frac{2}{3}. Then add the square of -\frac{2}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{4}{3}x+\frac{4}{9}=1+\frac{4}{9}
Square -\frac{2}{3} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{4}{3}x+\frac{4}{9}=\frac{13}{9}
Add 1 to \frac{4}{9}.
\left(x-\frac{2}{3}\right)^{2}=\frac{13}{9}
Factor x^{2}-\frac{4}{3}x+\frac{4}{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{2}{3}\right)^{2}}=\sqrt{\frac{13}{9}}
Take the square root of both sides of the equation.
x-\frac{2}{3}=\frac{\sqrt{13}}{3} x-\frac{2}{3}=-\frac{\sqrt{13}}{3}
Simplify.
x=\frac{\sqrt{13}+2}{3} x=\frac{2-\sqrt{13}}{3}
Add \frac{2}{3} to both sides of the equation.