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-3x^{2}-4x-5=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(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-3\right)\left(-5\right)}}{2\left(-3\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, -4 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-4\right)±\sqrt{16-4\left(-3\right)\left(-5\right)}}{2\left(-3\right)}
Square -4.
x=\frac{-\left(-4\right)±\sqrt{16+12\left(-5\right)}}{2\left(-3\right)}
Multiply -4 times -3.
x=\frac{-\left(-4\right)±\sqrt{16-60}}{2\left(-3\right)}
Multiply 12 times -5.
x=\frac{-\left(-4\right)±\sqrt{-44}}{2\left(-3\right)}
Add 16 to -60.
x=\frac{-\left(-4\right)±2\sqrt{11}i}{2\left(-3\right)}
Take the square root of -44.
x=\frac{4±2\sqrt{11}i}{2\left(-3\right)}
The opposite of -4 is 4.
x=\frac{4±2\sqrt{11}i}{-6}
Multiply 2 times -3.
x=\frac{4+2\sqrt{11}i}{-6}
Now solve the equation x=\frac{4±2\sqrt{11}i}{-6} when ± is plus. Add 4 to 2i\sqrt{11}.
x=\frac{-\sqrt{11}i-2}{3}
Divide 4+2i\sqrt{11} by -6.
x=\frac{-2\sqrt{11}i+4}{-6}
Now solve the equation x=\frac{4±2\sqrt{11}i}{-6} when ± is minus. Subtract 2i\sqrt{11} from 4.
x=\frac{-2+\sqrt{11}i}{3}
Divide 4-2i\sqrt{11} by -6.
x=\frac{-\sqrt{11}i-2}{3} x=\frac{-2+\sqrt{11}i}{3}
The equation is now solved.
-3x^{2}-4x-5=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.
-3x^{2}-4x-5-\left(-5\right)=-\left(-5\right)
Add 5 to both sides of the equation.
-3x^{2}-4x=-\left(-5\right)
Subtracting -5 from itself leaves 0.
-3x^{2}-4x=5
Subtract -5 from 0.
\frac{-3x^{2}-4x}{-3}=\frac{5}{-3}
Divide both sides by -3.
x^{2}+\left(-\frac{4}{-3}\right)x=\frac{5}{-3}
Dividing by -3 undoes the multiplication by -3.
x^{2}+\frac{4}{3}x=\frac{5}{-3}
Divide -4 by -3.
x^{2}+\frac{4}{3}x=-\frac{5}{3}
Divide 5 by -3.
x^{2}+\frac{4}{3}x+\left(\frac{2}{3}\right)^{2}=-\frac{5}{3}+\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}=-\frac{5}{3}+\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{11}{9}
Add -\frac{5}{3} to \frac{4}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{2}{3}\right)^{2}=-\frac{11}{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{11}{9}}
Take the square root of both sides of the equation.
x+\frac{2}{3}=\frac{\sqrt{11}i}{3} x+\frac{2}{3}=-\frac{\sqrt{11}i}{3}
Simplify.
x=\frac{-2+\sqrt{11}i}{3} x=\frac{-\sqrt{11}i-2}{3}
Subtract \frac{2}{3} from both sides of the equation.