Solve for x
x=-2
x=12
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Quadratic Equation
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-2x-4=- \frac{ 1 }{ 4 } { x }^{ 2 } + \frac{ 1 }{ 2 } x+2
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-2x-4+\frac{1}{4}x^{2}=\frac{1}{2}x+2
Add \frac{1}{4}x^{2} to both sides.
-2x-4+\frac{1}{4}x^{2}-\frac{1}{2}x=2
Subtract \frac{1}{2}x from both sides.
-\frac{5}{2}x-4+\frac{1}{4}x^{2}=2
Combine -2x and -\frac{1}{2}x to get -\frac{5}{2}x.
-\frac{5}{2}x-4+\frac{1}{4}x^{2}-2=0
Subtract 2 from both sides.
-\frac{5}{2}x-6+\frac{1}{4}x^{2}=0
Subtract 2 from -4 to get -6.
\frac{1}{4}x^{2}-\frac{5}{2}x-6=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(-\frac{5}{2}\right)±\sqrt{\left(-\frac{5}{2}\right)^{2}-4\times \frac{1}{4}\left(-6\right)}}{2\times \frac{1}{4}}
This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{4} for a, -\frac{5}{2} for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-\frac{5}{2}\right)±\sqrt{\frac{25}{4}-4\times \frac{1}{4}\left(-6\right)}}{2\times \frac{1}{4}}
Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.
x=\frac{-\left(-\frac{5}{2}\right)±\sqrt{\frac{25}{4}-\left(-6\right)}}{2\times \frac{1}{4}}
Multiply -4 times \frac{1}{4}.
x=\frac{-\left(-\frac{5}{2}\right)±\sqrt{\frac{25}{4}+6}}{2\times \frac{1}{4}}
Multiply -1 times -6.
x=\frac{-\left(-\frac{5}{2}\right)±\sqrt{\frac{49}{4}}}{2\times \frac{1}{4}}
Add \frac{25}{4} to 6.
x=\frac{-\left(-\frac{5}{2}\right)±\frac{7}{2}}{2\times \frac{1}{4}}
Take the square root of \frac{49}{4}.
x=\frac{\frac{5}{2}±\frac{7}{2}}{2\times \frac{1}{4}}
The opposite of -\frac{5}{2} is \frac{5}{2}.
x=\frac{\frac{5}{2}±\frac{7}{2}}{\frac{1}{2}}
Multiply 2 times \frac{1}{4}.
x=\frac{6}{\frac{1}{2}}
Now solve the equation x=\frac{\frac{5}{2}±\frac{7}{2}}{\frac{1}{2}} when ± is plus. Add \frac{5}{2} to \frac{7}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
x=12
Divide 6 by \frac{1}{2} by multiplying 6 by the reciprocal of \frac{1}{2}.
x=-\frac{1}{\frac{1}{2}}
Now solve the equation x=\frac{\frac{5}{2}±\frac{7}{2}}{\frac{1}{2}} when ± is minus. Subtract \frac{7}{2} from \frac{5}{2} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.
x=-2
Divide -1 by \frac{1}{2} by multiplying -1 by the reciprocal of \frac{1}{2}.
x=12 x=-2
The equation is now solved.
-2x-4+\frac{1}{4}x^{2}=\frac{1}{2}x+2
Add \frac{1}{4}x^{2} to both sides.
-2x-4+\frac{1}{4}x^{2}-\frac{1}{2}x=2
Subtract \frac{1}{2}x from both sides.
-\frac{5}{2}x-4+\frac{1}{4}x^{2}=2
Combine -2x and -\frac{1}{2}x to get -\frac{5}{2}x.
-\frac{5}{2}x+\frac{1}{4}x^{2}=2+4
Add 4 to both sides.
-\frac{5}{2}x+\frac{1}{4}x^{2}=6
Add 2 and 4 to get 6.
\frac{1}{4}x^{2}-\frac{5}{2}x=6
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{\frac{1}{4}x^{2}-\frac{5}{2}x}{\frac{1}{4}}=\frac{6}{\frac{1}{4}}
Multiply both sides by 4.
x^{2}+\left(-\frac{\frac{5}{2}}{\frac{1}{4}}\right)x=\frac{6}{\frac{1}{4}}
Dividing by \frac{1}{4} undoes the multiplication by \frac{1}{4}.
x^{2}-10x=\frac{6}{\frac{1}{4}}
Divide -\frac{5}{2} by \frac{1}{4} by multiplying -\frac{5}{2} by the reciprocal of \frac{1}{4}.
x^{2}-10x=24
Divide 6 by \frac{1}{4} by multiplying 6 by the reciprocal of \frac{1}{4}.
x^{2}-10x+\left(-5\right)^{2}=24+\left(-5\right)^{2}
Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-10x+25=24+25
Square -5.
x^{2}-10x+25=49
Add 24 to 25.
\left(x-5\right)^{2}=49
Factor x^{2}-10x+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(x-5\right)^{2}}=\sqrt{49}
Take the square root of both sides of the equation.
x-5=7 x-5=-7
Simplify.
x=12 x=-2
Add 5 to both sides of the equation.
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Simultaneous equation
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Integration
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Limits
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