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x^{2}-10x+25=3
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}-10x+25-3=3-3
Subtract 3 from both sides of the equation.
x^{2}-10x+25-3=0
Subtracting 3 from itself leaves 0.
x^{2}-10x+22=0
Subtract 3 from 25.
x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 22}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and 22 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-10\right)±\sqrt{100-4\times 22}}{2}
Square -10.
x=\frac{-\left(-10\right)±\sqrt{100-88}}{2}
Multiply -4 times 22.
x=\frac{-\left(-10\right)±\sqrt{12}}{2}
Add 100 to -88.
x=\frac{-\left(-10\right)±2\sqrt{3}}{2}
Take the square root of 12.
x=\frac{10±2\sqrt{3}}{2}
The opposite of -10 is 10.
x=\frac{2\sqrt{3}+10}{2}
Now solve the equation x=\frac{10±2\sqrt{3}}{2} when ± is plus. Add 10 to 2\sqrt{3}.
x=\sqrt{3}+5
Divide 10+2\sqrt{3} by 2.
x=\frac{10-2\sqrt{3}}{2}
Now solve the equation x=\frac{10±2\sqrt{3}}{2} when ± is minus. Subtract 2\sqrt{3} from 10.
x=5-\sqrt{3}
Divide 10-2\sqrt{3} by 2.
x=\sqrt{3}+5 x=5-\sqrt{3}
The equation is now solved.
x^{2}-10x+25=3
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.
\left(x-5\right)^{2}=3
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{3}
Take the square root of both sides of the equation.
x-5=\sqrt{3} x-5=-\sqrt{3}
Simplify.
x=\sqrt{3}+5 x=5-\sqrt{3}
Add 5 to both sides of the equation.