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6x^{2}-15x+40=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(-15\right)±\sqrt{\left(-15\right)^{2}-4\times 6\times 40}}{2\times 6}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -15 for b, and 40 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-15\right)±\sqrt{225-4\times 6\times 40}}{2\times 6}
Square -15.
x=\frac{-\left(-15\right)±\sqrt{225-24\times 40}}{2\times 6}
Multiply -4 times 6.
x=\frac{-\left(-15\right)±\sqrt{225-960}}{2\times 6}
Multiply -24 times 40.
x=\frac{-\left(-15\right)±\sqrt{-735}}{2\times 6}
Add 225 to -960.
x=\frac{-\left(-15\right)±7\sqrt{15}i}{2\times 6}
Take the square root of -735.
x=\frac{15±7\sqrt{15}i}{2\times 6}
The opposite of -15 is 15.
x=\frac{15±7\sqrt{15}i}{12}
Multiply 2 times 6.
x=\frac{15+7\sqrt{15}i}{12}
Now solve the equation x=\frac{15±7\sqrt{15}i}{12} when ± is plus. Add 15 to 7i\sqrt{15}.
x=\frac{7\sqrt{15}i}{12}+\frac{5}{4}
Divide 15+7i\sqrt{15} by 12.
x=\frac{-7\sqrt{15}i+15}{12}
Now solve the equation x=\frac{15±7\sqrt{15}i}{12} when ± is minus. Subtract 7i\sqrt{15} from 15.
x=-\frac{7\sqrt{15}i}{12}+\frac{5}{4}
Divide 15-7i\sqrt{15} by 12.
x=\frac{7\sqrt{15}i}{12}+\frac{5}{4} x=-\frac{7\sqrt{15}i}{12}+\frac{5}{4}
The equation is now solved.
6x^{2}-15x+40=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.
6x^{2}-15x+40-40=-40
Subtract 40 from both sides of the equation.
6x^{2}-15x=-40
Subtracting 40 from itself leaves 0.
\frac{6x^{2}-15x}{6}=-\frac{40}{6}
Divide both sides by 6.
x^{2}+\left(-\frac{15}{6}\right)x=-\frac{40}{6}
Dividing by 6 undoes the multiplication by 6.
x^{2}-\frac{5}{2}x=-\frac{40}{6}
Reduce the fraction \frac{-15}{6} to lowest terms by extracting and canceling out 3.
x^{2}-\frac{5}{2}x=-\frac{20}{3}
Reduce the fraction \frac{-40}{6} to lowest terms by extracting and canceling out 2.
x^{2}-\frac{5}{2}x+\left(-\frac{5}{4}\right)^{2}=-\frac{20}{3}+\left(-\frac{5}{4}\right)^{2}
Divide -\frac{5}{2}, the coefficient of the x term, by 2 to get -\frac{5}{4}. Then add the square of -\frac{5}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{5}{2}x+\frac{25}{16}=-\frac{20}{3}+\frac{25}{16}
Square -\frac{5}{4} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{5}{2}x+\frac{25}{16}=-\frac{245}{48}
Add -\frac{20}{3} to \frac{25}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{5}{4}\right)^{2}=-\frac{245}{48}
Factor x^{2}-\frac{5}{2}x+\frac{25}{16}. 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}{4}\right)^{2}}=\sqrt{-\frac{245}{48}}
Take the square root of both sides of the equation.
x-\frac{5}{4}=\frac{7\sqrt{15}i}{12} x-\frac{5}{4}=-\frac{7\sqrt{15}i}{12}
Simplify.
x=\frac{7\sqrt{15}i}{12}+\frac{5}{4} x=-\frac{7\sqrt{15}i}{12}+\frac{5}{4}
Add \frac{5}{4} to both sides of the equation.