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x^{2}+2x-\frac{5}{4}=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{-2±\sqrt{2^{2}-4\left(-\frac{5}{4}\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 2 for b, and -\frac{5}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-2±\sqrt{4-4\left(-\frac{5}{4}\right)}}{2}
Square 2.
x=\frac{-2±\sqrt{4+5}}{2}
Multiply -4 times -\frac{5}{4}.
x=\frac{-2±\sqrt{9}}{2}
Add 4 to 5.
x=\frac{-2±3}{2}
Take the square root of 9.
x=\frac{1}{2}
Now solve the equation x=\frac{-2±3}{2} when ± is plus. Add -2 to 3.
x=-\frac{5}{2}
Now solve the equation x=\frac{-2±3}{2} when ± is minus. Subtract 3 from -2.
x=\frac{1}{2} x=-\frac{5}{2}
The equation is now solved.
x^{2}+2x-\frac{5}{4}=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.
x^{2}+2x-\frac{5}{4}-\left(-\frac{5}{4}\right)=-\left(-\frac{5}{4}\right)
Add \frac{5}{4} to both sides of the equation.
x^{2}+2x=-\left(-\frac{5}{4}\right)
Subtracting -\frac{5}{4} from itself leaves 0.
x^{2}+2x=\frac{5}{4}
Subtract -\frac{5}{4} from 0.
x^{2}+2x+1^{2}=\frac{5}{4}+1^{2}
Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+2x+1=\frac{5}{4}+1
Square 1.
x^{2}+2x+1=\frac{9}{4}
Add \frac{5}{4} to 1.
\left(x+1\right)^{2}=\frac{9}{4}
Factor x^{2}+2x+1. 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+1\right)^{2}}=\sqrt{\frac{9}{4}}
Take the square root of both sides of the equation.
x+1=\frac{3}{2} x+1=-\frac{3}{2}
Simplify.
x=\frac{1}{2} x=-\frac{5}{2}
Subtract 1 from both sides of the equation.