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x^{2}+x-4=-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}+x-4-\left(-1\right)=-1-\left(-1\right)
Add 1 to both sides of the equation.
x^{2}+x-4-\left(-1\right)=0
Subtracting -1 from itself leaves 0.
x^{2}+x-3=0
Subtract -1 from -4.
x=\frac{-1±\sqrt{1^{2}-4\left(-3\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 1 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-1±\sqrt{1-4\left(-3\right)}}{2}
Square 1.
x=\frac{-1±\sqrt{1+12}}{2}
Multiply -4 times -3.
x=\frac{-1±\sqrt{13}}{2}
Add 1 to 12.
x=\frac{\sqrt{13}-1}{2}
Now solve the equation x=\frac{-1±\sqrt{13}}{2} when ± is plus. Add -1 to \sqrt{13}.
x=\frac{-\sqrt{13}-1}{2}
Now solve the equation x=\frac{-1±\sqrt{13}}{2} when ± is minus. Subtract \sqrt{13} from -1.
x=\frac{\sqrt{13}-1}{2} x=\frac{-\sqrt{13}-1}{2}
The equation is now solved.
x^{2}+x-4=-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}+x-4-\left(-4\right)=-1-\left(-4\right)
Add 4 to both sides of the equation.
x^{2}+x=-1-\left(-4\right)
Subtracting -4 from itself leaves 0.
x^{2}+x=3
Subtract -4 from -1.
x^{2}+x+\left(\frac{1}{2}\right)^{2}=3+\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}=3+\frac{1}{4}
Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}+x+\frac{1}{4}=\frac{13}{4}
Add 3 to \frac{1}{4}.
\left(x+\frac{1}{2}\right)^{2}=\frac{13}{4}
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{\frac{13}{4}}
Take the square root of both sides of the equation.
x+\frac{1}{2}=\frac{\sqrt{13}}{2} x+\frac{1}{2}=-\frac{\sqrt{13}}{2}
Simplify.
x=\frac{\sqrt{13}-1}{2} x=\frac{-\sqrt{13}-1}{2}
Subtract \frac{1}{2} from both sides of the equation.