Solve for x
x=-\frac{6}{7}\approx -0.857142857
x=0
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x\left(-7x-6\right)=0
Factor out x.
x=0 x=-\frac{6}{7}
To find equation solutions, solve x=0 and -7x-6=0.
-7x^{2}-6x=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(-6\right)±\sqrt{\left(-6\right)^{2}}}{2\left(-7\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -7 for a, -6 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-6\right)±6}{2\left(-7\right)}
Take the square root of \left(-6\right)^{2}.
x=\frac{6±6}{2\left(-7\right)}
The opposite of -6 is 6.
x=\frac{6±6}{-14}
Multiply 2 times -7.
x=\frac{12}{-14}
Now solve the equation x=\frac{6±6}{-14} when ± is plus. Add 6 to 6.
x=-\frac{6}{7}
Reduce the fraction \frac{12}{-14} to lowest terms by extracting and canceling out 2.
x=\frac{0}{-14}
Now solve the equation x=\frac{6±6}{-14} when ± is minus. Subtract 6 from 6.
x=0
Divide 0 by -14.
x=-\frac{6}{7} x=0
The equation is now solved.
-7x^{2}-6x=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{-7x^{2}-6x}{-7}=\frac{0}{-7}
Divide both sides by -7.
x^{2}+\left(-\frac{6}{-7}\right)x=\frac{0}{-7}
Dividing by -7 undoes the multiplication by -7.
x^{2}+\frac{6}{7}x=\frac{0}{-7}
Divide -6 by -7.
x^{2}+\frac{6}{7}x=0
Divide 0 by -7.
x^{2}+\frac{6}{7}x+\left(\frac{3}{7}\right)^{2}=\left(\frac{3}{7}\right)^{2}
Divide \frac{6}{7}, the coefficient of the x term, by 2 to get \frac{3}{7}. Then add the square of \frac{3}{7} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{6}{7}x+\frac{9}{49}=\frac{9}{49}
Square \frac{3}{7} by squaring both the numerator and the denominator of the fraction.
\left(x+\frac{3}{7}\right)^{2}=\frac{9}{49}
Factor x^{2}+\frac{6}{7}x+\frac{9}{49}. 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{3}{7}\right)^{2}}=\sqrt{\frac{9}{49}}
Take the square root of both sides of the equation.
x+\frac{3}{7}=\frac{3}{7} x+\frac{3}{7}=-\frac{3}{7}
Simplify.
x=0 x=-\frac{6}{7}
Subtract \frac{3}{7} from both sides of the equation.
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Limits
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