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x^{2}-4x+5-5=0
Subtract 5 from both sides.
x^{2}-4x=0
Subtract 5 from 5 to get 0.
x\left(x-4\right)=0
Factor out x.
x=0 x=4
To find equation solutions, solve x=0 and x-4=0.
x^{2}-4x+5=5
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^{2}-4x+5-5=5-5
Subtract 5 from both sides of the equation.
x^{2}-4x+5-5=0
Subtracting 5 from itself leaves 0.
x^{2}-4x=0
Subtract 5 from 5.
x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -4 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-4\right)±4}{2}
Take the square root of \left(-4\right)^{2}.
x=\frac{4±4}{2}
The opposite of -4 is 4.
x=\frac{8}{2}
Now solve the equation x=\frac{4±4}{2} when ± is plus. Add 4 to 4.
x=4
Divide 8 by 2.
x=\frac{0}{2}
Now solve the equation x=\frac{4±4}{2} when ± is minus. Subtract 4 from 4.
x=0
Divide 0 by 2.
x=4 x=0
The equation is now solved.
x^{2}-4x+5-5=0
Subtract 5 from both sides.
x^{2}-4x=0
Subtract 5 from 5 to get 0.
x^{2}-4x+\left(-2\right)^{2}=\left(-2\right)^{2}
Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-4x+4=4
Square -2.
\left(x-2\right)^{2}=4
Factor x^{2}-4x+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-2\right)^{2}}=\sqrt{4}
Take the square root of both sides of the equation.
x-2=2 x-2=-2
Simplify.
x=4 x=0
Add 2 to both sides of the equation.