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4x^{2}-4x-19=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\times 4\left(-19\right)}}{2\times 4}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -4 for b, and -19 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-4\right)±\sqrt{16-4\times 4\left(-19\right)}}{2\times 4}
Square -4.
x=\frac{-\left(-4\right)±\sqrt{16-16\left(-19\right)}}{2\times 4}
Multiply -4 times 4.
x=\frac{-\left(-4\right)±\sqrt{16+304}}{2\times 4}
Multiply -16 times -19.
x=\frac{-\left(-4\right)±\sqrt{320}}{2\times 4}
Add 16 to 304.
x=\frac{-\left(-4\right)±8\sqrt{5}}{2\times 4}
Take the square root of 320.
x=\frac{4±8\sqrt{5}}{2\times 4}
The opposite of -4 is 4.
x=\frac{4±8\sqrt{5}}{8}
Multiply 2 times 4.
x=\frac{8\sqrt{5}+4}{8}
Now solve the equation x=\frac{4±8\sqrt{5}}{8} when ± is plus. Add 4 to 8\sqrt{5}.
x=\sqrt{5}+\frac{1}{2}
Divide 4+8\sqrt{5} by 8.
x=\frac{4-8\sqrt{5}}{8}
Now solve the equation x=\frac{4±8\sqrt{5}}{8} when ± is minus. Subtract 8\sqrt{5} from 4.
x=\frac{1}{2}-\sqrt{5}
Divide 4-8\sqrt{5} by 8.
x=\sqrt{5}+\frac{1}{2} x=\frac{1}{2}-\sqrt{5}
The equation is now solved.
4x^{2}-4x-19=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.
4x^{2}-4x-19-\left(-19\right)=-\left(-19\right)
Add 19 to both sides of the equation.
4x^{2}-4x=-\left(-19\right)
Subtracting -19 from itself leaves 0.
4x^{2}-4x=19
Subtract -19 from 0.
\frac{4x^{2}-4x}{4}=\frac{19}{4}
Divide both sides by 4.
x^{2}+\left(-\frac{4}{4}\right)x=\frac{19}{4}
Dividing by 4 undoes the multiplication by 4.
x^{2}-x=\frac{19}{4}
Divide -4 by 4.
x^{2}-x+\left(-\frac{1}{2}\right)^{2}=\frac{19}{4}+\left(-\frac{1}{2}\right)^{2}
Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-x+\frac{1}{4}=\frac{19+1}{4}
Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}-x+\frac{1}{4}=5
Add \frac{19}{4} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{1}{2}\right)^{2}=5
Factor x^{2}-x+\frac{1}{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{1}{2}\right)^{2}}=\sqrt{5}
Take the square root of both sides of the equation.
x-\frac{1}{2}=\sqrt{5} x-\frac{1}{2}=-\sqrt{5}
Simplify.
x=\sqrt{5}+\frac{1}{2} x=\frac{1}{2}-\sqrt{5}
Add \frac{1}{2} to both sides of the equation.