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x^{2}-13x+42=1
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}-13x+42-1=1-1
Subtract 1 from both sides of the equation.
x^{2}-13x+42-1=0
Subtracting 1 from itself leaves 0.
x^{2}-13x+41=0
Subtract 1 from 42.
x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 41}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -13 for b, and 41 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-13\right)±\sqrt{169-4\times 41}}{2}
Square -13.
x=\frac{-\left(-13\right)±\sqrt{169-164}}{2}
Multiply -4 times 41.
x=\frac{-\left(-13\right)±\sqrt{5}}{2}
Add 169 to -164.
x=\frac{13±\sqrt{5}}{2}
The opposite of -13 is 13.
x=\frac{\sqrt{5}+13}{2}
Now solve the equation x=\frac{13±\sqrt{5}}{2} when ± is plus. Add 13 to \sqrt{5}.
x=\frac{13-\sqrt{5}}{2}
Now solve the equation x=\frac{13±\sqrt{5}}{2} when ± is minus. Subtract \sqrt{5} from 13.
x=\frac{\sqrt{5}+13}{2} x=\frac{13-\sqrt{5}}{2}
The equation is now solved.
x^{2}-13x+42=1
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.
x^{2}-13x+42-42=1-42
Subtract 42 from both sides of the equation.
x^{2}-13x=1-42
Subtracting 42 from itself leaves 0.
x^{2}-13x=-41
Subtract 42 from 1.
x^{2}-13x+\left(-\frac{13}{2}\right)^{2}=-41+\left(-\frac{13}{2}\right)^{2}
Divide -13, the coefficient of the x term, by 2 to get -\frac{13}{2}. Then add the square of -\frac{13}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-13x+\frac{169}{4}=-41+\frac{169}{4}
Square -\frac{13}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}-13x+\frac{169}{4}=\frac{5}{4}
Add -41 to \frac{169}{4}.
\left(x-\frac{13}{2}\right)^{2}=\frac{5}{4}
Factor x^{2}-13x+\frac{169}{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{13}{2}\right)^{2}}=\sqrt{\frac{5}{4}}
Take the square root of both sides of the equation.
x-\frac{13}{2}=\frac{\sqrt{5}}{2} x-\frac{13}{2}=-\frac{\sqrt{5}}{2}
Simplify.
x=\frac{\sqrt{5}+13}{2} x=\frac{13-\sqrt{5}}{2}
Add \frac{13}{2} to both sides of the equation.