Solve for c
c=\frac{2}{5}=0.4
c=0
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c\left(5c-2\right)=0
Factor out c.
c=0 c=\frac{2}{5}
To find equation solutions, solve c=0 and 5c-2=0.
5c^{2}-2c=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.
c=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}}}{2\times 5}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -2 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
c=\frac{-\left(-2\right)±2}{2\times 5}
Take the square root of \left(-2\right)^{2}.
c=\frac{2±2}{2\times 5}
The opposite of -2 is 2.
c=\frac{2±2}{10}
Multiply 2 times 5.
c=\frac{4}{10}
Now solve the equation c=\frac{2±2}{10} when ± is plus. Add 2 to 2.
c=\frac{2}{5}
Reduce the fraction \frac{4}{10} to lowest terms by extracting and canceling out 2.
c=\frac{0}{10}
Now solve the equation c=\frac{2±2}{10} when ± is minus. Subtract 2 from 2.
c=0
Divide 0 by 10.
c=\frac{2}{5} c=0
The equation is now solved.
5c^{2}-2c=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{5c^{2}-2c}{5}=\frac{0}{5}
Divide both sides by 5.
c^{2}-\frac{2}{5}c=\frac{0}{5}
Dividing by 5 undoes the multiplication by 5.
c^{2}-\frac{2}{5}c=0
Divide 0 by 5.
c^{2}-\frac{2}{5}c+\left(-\frac{1}{5}\right)^{2}=\left(-\frac{1}{5}\right)^{2}
Divide -\frac{2}{5}, the coefficient of the x term, by 2 to get -\frac{1}{5}. Then add the square of -\frac{1}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
c^{2}-\frac{2}{5}c+\frac{1}{25}=\frac{1}{25}
Square -\frac{1}{5} by squaring both the numerator and the denominator of the fraction.
\left(c-\frac{1}{5}\right)^{2}=\frac{1}{25}
Factor c^{2}-\frac{2}{5}c+\frac{1}{25}. 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(c-\frac{1}{5}\right)^{2}}=\sqrt{\frac{1}{25}}
Take the square root of both sides of the equation.
c-\frac{1}{5}=\frac{1}{5} c-\frac{1}{5}=-\frac{1}{5}
Simplify.
c=\frac{2}{5} c=0
Add \frac{1}{5} to both sides of the equation.
Examples
Quadratic equation
{ x } ^ { 2 } - 4 x - 5 = 0
Trigonometry
4 \sin \theta \cos \theta = 2 \sin \theta
Linear equation
y = 3x + 4
Arithmetic
699 * 533
Matrix
\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
Limits
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}