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x\left(2x+14\right)=0
Factor out x.
x=0 x=-7
To find equation solutions, solve x=0 and 2x+14=0.
2x^{2}+14x=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{-14±\sqrt{14^{2}}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 14 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-14±14}{2\times 2}
Take the square root of 14^{2}.
x=\frac{-14±14}{4}
Multiply 2 times 2.
x=\frac{0}{4}
Now solve the equation x=\frac{-14±14}{4} when ± is plus. Add -14 to 14.
x=0
Divide 0 by 4.
x=-\frac{28}{4}
Now solve the equation x=\frac{-14±14}{4} when ± is minus. Subtract 14 from -14.
x=-7
Divide -28 by 4.
x=0 x=-7
The equation is now solved.
2x^{2}+14x=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{2x^{2}+14x}{2}=\frac{0}{2}
Divide both sides by 2.
x^{2}+\frac{14}{2}x=\frac{0}{2}
Dividing by 2 undoes the multiplication by 2.
x^{2}+7x=\frac{0}{2}
Divide 14 by 2.
x^{2}+7x=0
Divide 0 by 2.
x^{2}+7x+\left(\frac{7}{2}\right)^{2}=\left(\frac{7}{2}\right)^{2}
Divide 7, the coefficient of the x term, by 2 to get \frac{7}{2}. Then add the square of \frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+7x+\frac{49}{4}=\frac{49}{4}
Square \frac{7}{2} by squaring both the numerator and the denominator of the fraction.
\left(x+\frac{7}{2}\right)^{2}=\frac{49}{4}
Factor x^{2}+7x+\frac{49}{4}. 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{7}{2}\right)^{2}}=\sqrt{\frac{49}{4}}
Take the square root of both sides of the equation.
x+\frac{7}{2}=\frac{7}{2} x+\frac{7}{2}=-\frac{7}{2}
Simplify.
x=0 x=-7
Subtract \frac{7}{2} from both sides of the equation.